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A131191
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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.
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2
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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
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OFFSET
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1,1
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COMMENTS
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If n is in this sequence, then so is n+121. - Max Alekseyev, Feb 02 2015
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LINKS
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FORMULA
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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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Initial terms were calculated by Peter J. C. Moses; see comment in A129995.
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STATUS
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approved
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