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A343132
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a(n) is the quotient obtained when integer A343131(n) = k is divided by A061486(k).
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1
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1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 10, 1, 10, 1, 10, 3, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 100, 6, 5, 24, 11, 10, 1, 100, 9, 8, 42, 13, 13, 10, 1, 100, 43, 16, 22, 10, 1, 100, 30, 9, 2, 10, 1, 100, 4, 3, 10, 1, 100, 31, 6, 5, 10, 1, 100, 15, 10, 1, 100, 13, 11, 10, 1, 100, 10, 1, 1000
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OFFSET
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1,10
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COMMENTS
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The first 9 terms corresponding to the 1-digit numbers k = u are the quotients u/u = 1.
The next 19 terms from a(10) = 10 to a(28) = 1 corresponding to 2-digit numbers k = du are the quotients du/(d+u + d*u).
The next 50 terms from a(29) = 100 to a(78) = 1 corresponding to 3-digit numbers k = hdu (in A328864) are the quotients hdu/f_3(h,d,u) where f_3(h,d,u) = (h+d+u) + (h*d+d*u+u*h) + (h*d*u).
The next 87 terms, from a(79) = 1000 to a(165) = 1, corresponding to 4-digit numbers k = thdu are the quotients thdu/f_4(t,h,d,u) where f_4(t,h,d,u) = (t+h+d+u) + (t*h+t*d+t*u+h*d+h*u+d*u) + (t*h*d+t*h*u+t*d*u+h*d*u) + (t*h*d*u).
When A343131(n) = z*10^q = A037124(r) is a number that contains only one nonzero digit z, then A061486(A037124(r)) = this nonzero digit z and a(n) = 10^q.
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LINKS
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FORMULA
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EXAMPLE
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For A343131(17) = 42, A061486(42) = 4+2 + 4*2 = 14 and a(17) = 42/14 = 3.
For A343131(58) = 573, A061486(573) = 5+7+3 + 5*7+7*3+3*5 + 5*7*3 = 191 and a(58) = 573/191 = 3.
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PROG
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(PARI) sympol(X, n) = my(s=0); forvec(i=vector(n, j, [1, #X]), s+=prod(k=1, n, X[i[k]]), 2); s ;
f(n) = my(d=digits(n)); sum(k=1, #d, sympol(d, k));
lista(nn) = {for (n=1, nn, my(q = n/f(n)); if (denominator(q) == 1, print1(q, ", ")); ); } \\ Michel Marcus, Apr 08 2021
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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