A131191 Numbers n>=0 such that d(n) = (n^1 + 1) (n^2 + 2) ... (n^22 + 22) / 22!, e(n) = (n^1 + 1) (n^2 + 2) ... (n^23 + 23) / 23!, and f(n) = (n^1 + 1) (n^2 + 2) ... (n^24 + 24) / 24! take nonintegral values.

7, 18, 29, 40, 51, 62, 73, 84, 95, 106, 128, 139, 150, 161, 172, 183, 194, 205, 216, 227, 249, 260, 271, 282, 293, 304, 315, 326, 337, 348, 370, 381, 392, 403, 414, 425, 436, 447, 458, 469, 491, 502, 513, 524, 535, 546, 557, 568, 579, 590, 612, 623, 634, 645, 656, 667, 678, 689, 700, 711, 733, 744

Offset: 1

Author

Alexander R. Povolotsky, Sep 25 2007

Keywords

• nonn

Comments

If n is in this sequence, then so is n+121. - Max Alekseyev, Feb 02 2015

Crossrefs

Cf. A017473, A129995

Formula

Notice that 22! = 2^19 * 3^9 * 5^4 * 7^3 * 11^2 * 13 * 17 * 19. All these prime powers divide (n^1 + 1)*(n^2 + 2)*(n^3 +3)*...*(n^22 + 22), except for 11^2. 11^2 does not divide (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^22 + 22) for n = 7, 18, 29, 40, 51, 62, 73, 84, 95, 106 modulo 121. That is, d(n) is nonintegral for n the form 11m+7 but not 121m+117, and so are e(n) and f(n). - Max Alekseyev, Nov 10 2007

Extensions

Initial terms were calculated by Peter J. C. Moses; see comment in A129995.

More terms from Max Alekseyev, Feb 02 2015