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A309745
Indices k of highly composite numbers with records of low values of the ratio between consecutive terms, A002182(k+1)/A002182(k).
0
1, 3, 7, 8, 14, 24, 37, 65, 97, 105, 145, 163, 253, 686, 1061, 1871, 2025, 15255, 28092, 36183, 56485, 81294, 81993, 173338, 328432, 557890
OFFSET
1,2
COMMENTS
Ramanujan proved that the asymptotic limit of the ratio between consecutive highly composite numbers is 1. Therefore this sequence is infinite.
The first 26 terms were calculated from Achim Flammenkamp's list of the first 779674 highly composite numbers.
The corresponding highly composite numbers are A002182(a(n)) = 1, 4, 36, 48, 720, 25200, 665280, 698377680, 1606268664000, 8995104518400, 72779390658374400, ... and their corresponding consecutive terms are A002182(a(n)+1) = 2, 6, 48, 60, 840, 27720, 720720, 735134400, 1686582097200, 9316358251200, 74801040398884800, ...
The corresponding record ratios for the first 20 terms are of the form 1 + 1/m with m being an integer. The list of values of m is 1, 2, 3, 4, 6, 10, 12, 19, 20, 28, 36, 41, 176, 254, 345, 812, 9338, 10366, 21339, 44084, 89733/2, 497845/2, 435046, 800355, 30857708/23, 18882356170/7757, ...
LINKS
Achim Flammenkamp, Highly Composite Numbers.
Srinivasa Ramanujan, Highly composite numbers, Proceedings of the London Mathematical Society, Series 2, Vol. 14, No. 1 (1915), pp. 347-409, alternative link.
EXAMPLE
The first 3 terms of the sequence are 1, 3, 7. A002182(1+1)/A002182(1) = 2/1 = 2, A002182(3+1)/A002182(3) = 6/4 = 3/2, A002182(7+1)/A002182(7) = 48/36 = 4/3, ... and 2 > 3/2 > 4/3 > ...
MATHEMATICA
s={}; hcn1 = 1; dm = 1; rm = 3; c=0; Do[d = DivisorSigma[0, n]; If[d > dm, dm = d; hcn2 = n; c++; r = hcn2/hcn1; If[r < rm, rm = r; AppendTo[s, c]]; hcn1 = hcn2], {n, 2, 10^6}]; s
CROSSREFS
Cf. A002182.
Sequence in context: A127441 A067064 A093722 * A002381 A131559 A366783
KEYWORD
nonn,more
AUTHOR
Amiram Eldar, Aug 15 2019
STATUS
approved