A131992 a(n) = 1 + prime(n) + prime(n)^2 + prime(n)^3 + prime(n)^4.

31, 121, 781, 2801, 16105, 30941, 88741, 137561, 292561, 732541, 954305, 1926221, 2896405, 3500201, 4985761, 8042221, 12326281, 14076605, 20456441, 25774705, 28792661, 39449441, 48037081, 63455221, 89451461, 105101005, 113654321

Offset

1,1

Comments

Thébault shows that a(2) = 121 is the only square in this sequence. - Charles R Greathouse IV, Jul 23 2013

Giovanni Resta has found that 28792661 is the first Sophie Germain prime of this form (and actually of the form p = (n^m-1)/(n-1) for any p-1 > n, m > 1). - M. F. Hasler, Mar 03 2020

References

Victor Thébault, Curiosités arithmétiques, Mathesis 62 (1953), pp. 120-129.

Links

Ivan Panchenko, Table of n, a(n) for n = 1..1000

Formula

a(n) = 1 + A131991(n)*A000040(n).

a(n) = (A050997(n) - 1)/A006093(n).

a(n) = A000203(prime(n)^4). - R. J. Mathar, Mar 15 2018

a(n) = (prime(n)^5 - 1)/(prime(n) - 1) = A053699(prime(n)). (This is also meant by the 2nd formula.) - M. F. Hasler, Mar 03 2020

Example

a(1) = 31 because prime(1) = 2 and 1 + 2 + 2^2 + 2^3 + 2^4 = 1 + 2 + 4 + 8 + 16 = 31.

Mathematica

Table[Sum[Prime[n]^k, {k, 0, 4}], {n, 30}] (* Alonso del Arte, May 24 2015 *)

Prog

(PARI) a(n)=sigma(prime(n)^4)  \\ Charles R Greathouse IV, Jul 23 2013

Crossrefs

Cf. A030514, A008864, A060800, A131993.

Equals A053699 restricted to prime indices. Subsequence of primes is A190527.

Sequence in context: A202994 A038992 A068021 * A042884 A153073 A042886

Adjacent sequences: A131989 A131990 A131991 * A131993 A131994 A131995

Keyword

nonn,easy

Author

Reinhard Zumkeller, Aug 06 2007

Status

approved