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A116184
Numbers n such that 37^3 divides the numerator of generalized harmonic number H(36,n) = Sum[ 1/k^n, {k,1,36} ].
0
3, 37, 39, 73, 75, 111, 147, 148, 183, 185, 219, 221, 255, 259, 291, 295, 327, 333, 363, 369, 399, 407, 435, 443, 471, 481, 507, 517, 543, 555, 579, 591, 615, 629, 651, 665, 687, 703, 723, 739, 759, 777, 795, 813, 831, 851, 867, 887, 903, 925, 939, 961, 975
OFFSET
1,1
COMMENTS
Note the pattern in the first differences of a(n): {34,2,34,2,36,36,1,35,2,34,2,34,4,32,4,32,6,30,6,30,8,28,8,28,10,26,10,26,12,24,12,24,14,22,14,22,16,20,16,20,18,18,18,18,20,16,20,16,22,14,22,14,24,...}. Conjecture: All terms of the arithmetic progression 3+36k belong to a(n). Prime terms in a(n) are {3, 37, 73, 443, 739, 887, 1109, ...}. It appears that all primes in a(n) that are greater than 37 are of the form 37k-1. For example, 73 = 37*2-1, 443 = 37*12-1, 739 = 37*20-1, 887 = 37*24-1, 1109 = 37*30-1. Many terms in a(n) are the multiples of 37. There are terms of the form 37*m with m = {1,3,4,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,37,39,41,...}. Note that 37^4 divides the numerator of generalized harmonic number H(36,n) for n = {111, 147, 1047, 1369, 1443, 1479, ...} = {3*37, 3+4*36, 3+29*36, 37^2, 3+40*36, 3+41*36, ...}.
LINKS
Eric Weisstein, The World of Mathematics: Wolstenholme's Theorem.
Eric Weisstein, The World of Mathematics: Harmonic Number.
MATHEMATICA
Do[ f=Numerator[ Sum[ 1/k^n, {k, 1, 36} ] ]; If[ IntegerQ[ f/37^3 ], Print[n] ], {n, 1, 1000}]
CROSSREFS
Cf. A007408 = Wolstenholme numbers: numerator of Sum_{k=1..n} 1/k^3. Cf. A119722, A017533.
Sequence in context: A140448 A128061 A176240 * A037000 A042333 A106103
KEYWORD
hard,nonn
AUTHOR
Alexander Adamchuk, Apr 08 2007
STATUS
approved