A131190 Numbers n>=0 such that d(n) = (n^1 + 1) (n^2 + 2) ... (n^25 + 25) / 25! is nonintegral.
2, 7, 12, 18, 22, 27, 29, 37, 40, 47, 51, 52, 62, 72, 73, 77, 84, 87, 95, 97, 102, 106, 112, 122, 127, 128, 137, 139, 147, 150, 152, 161, 162, 172, 177, 183, 187, 194, 197, 202, 205, 212, 216, 222, 227, 237, 247, 249, 252, 260, 262, 271, 272, 277, 282, 287, 293, 297, 302, 304, 312, 315, 322, 326, 327, 337
Offset: 1
Keywords
Programs
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PARI
{ is_A131190(n) = setsearch([2,12,22,27,37,47],n%50) || ( (n%11)==7 && (n%121)!=117 ) } /* Max Alekseyev, Feb 02 2015 */
Formula
Notice that 25! = 2^22 * 3^10 * 5^6 * 7^3 * 11^2 * 13 * 17 * 19 * 23. The value of (n^1+1)(n^2+2)...(n^25+25) is always divisible by all these prime powers, except 5^6 and 11^2. There is no divisibility by 5^6 for n in {50m+2, 50m+12, 50m+22, 50m+27, 50m+37, 50m+47} and by 11^2 for n in {11m+7} \ {121m+117}. Therefore, the sequence is the union {50m+2} U {50m+12} U {50m+22} U {50m+27} U {50m+37} U {50m+47} U ( {11m+7} \ {121m+117} ). - 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
Comments