From 294d76333e8514c4e86aa1e40c6b9629d0e0ef5b Mon Sep 17 00:00:00 2001
From: Nathan Wagner <nw@hydaspes.if.org>
Date: Sat, 11 Mar 2017 03:49:31 -0600
Subject: [PATCH] added program to find consecutive partitions

---
 conseqpart.c | 67 ++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 67 insertions(+)
 create mode 100644 conseqpart.c

diff --git a/conseqpart.c b/conseqpart.c
new file mode 100644
index 0000000..8d5904c
--- /dev/null
+++ b/conseqpart.c
@@ -0,0 +1,67 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <math.h>
+
+/*
+ * observations:
+ *
+ * solutions of any length L starting at b have the sum
+ * S = L * (L-1 + 2b ) / 2 
+ *
+ * This sum must equal the target t, so the base for a given length is
+ * t = L * (L-1+2b) / 2
+ * b = (2t/L-L+1)/2
+ *
+ * Since b must be an integer, t % ( (L+1)*L / 2 ) must equal 0
+ * 2t % L must be zero and
+ * 2t/L - L + 1 must be even
+ *
+ * With a minimum base of 1, the sum is (L+1)*L / 2, 
+ * which must not exceed the target
+ * t >= (L+1)*L/2
+ * L <= (sqrt(8t+1)-1) / 2
+ *
+ * Even targets must have solution lengths >= 3
+ * even length solutions can start on any number
+ * odd length solutions must be centered on an even number
+ *
+ * Odd targets must have an odd number of odd numbers
+ * in the solution
+ * they always have an even length solution of t/2 + t/2 + 1
+ *
+ */
+int main(int ac, char *av[]) {
+	unsigned target; /* number to decompose */
+	unsigned sbase; /* lowest number in solution */
+	unsigned slen; /*  N quantity of numbers in solution */
+	int oddtarget;
+	unsigned maxlen;
+
+	target = atoi(av[1]);
+
+	maxlen = (sqrt(8.0*target+1.0)-1) / 2;
+	oddtarget = target % 2;
+
+	if (oddtarget) {
+		printf("2 %d .. %d\n", target/2, target/2+1);
+	}
+
+	for (slen=3; slen <= maxlen; slen++) {
+		if (oddtarget && (slen % 2 == 0 && (slen/2 % 2 == 0))) {
+			continue;
+		}
+
+		if (2 * target % slen != 0) {
+			continue;
+		}
+		if ( (2 * target / slen - slen + 1) % 2 != 0) {
+			continue;
+		}
+ 
+		sbase = (2 * target / slen - slen + 1) / 2;
+ 	
+		printf("%d %d .. %d\n", slen, sbase, sbase+slen-1);
+	}
+
+	return 0;
+}
-- 
2.40.0