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A324214
Sequence lists numbers k > 1 such that k^2 == phi(k) (mod sigma(k)), where phi = A000010 and sigma = A000203.
3
2, 4, 8, 12, 16, 32, 64, 76, 114, 128, 170, 256, 512, 1024, 1824, 2048, 2166, 4096, 7752, 8192, 16384, 16514, 28896, 32768, 41154, 59400, 65536, 68526, 90914, 131072, 177714, 230280, 262144, 276002, 428064, 524288, 722400, 781926, 1048576, 1299674, 1414272, 1546488
OFFSET
1,1
COMMENTS
All powers of 2 (A000079) are part of the sequence.
Yes, proof: if k = 2^q with q >= 1, we have: k^2 = 2^(2q), phi(k) = 2^(q-1) and sigma(k) = 2^(q+1) - 1. Then, k^2 - phi(k) = 2^(q-1) * sigma(k). - Bernard Schott, Feb 20 2019
Contains 6*19^k for k >= 1. - David A. Corneth, May 29 2019
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..180 (first 80 terms from Robert Israel)
FORMULA
Solutions of k^2 mod sigma(k) = phi(k).
EXAMPLE
sigma(114) = 240 and 114^2 mod 240 = 36 = phi(114).
MAPLE
with(numtheory): op(select(n->n^2 mod sigma(n)=phi(n), [$1..1546488]));
MATHEMATICA
Select[Range[10^5], Mod[#^2, DivisorSigma[1, #]] == EulerPhi[#] &] (* Giovanni Resta, May 29 2019 *)
PROG
(PARI) isok(k) = (k > 1) && ((k^2 % sigma(k)) == eulerphi(k)); \\ Michel Marcus, Feb 20 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Paolo P. Lava, Feb 18 2019
STATUS
approved