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A343131
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For m >= 1, the m-digit number k = d_m||...||d_2||d_1 is a term if it is divisible by f_m(k) that is the sum of the m elementary symmetric polynomials in m variables e_i(k): f_m(k) = Sum_{i=1..m} e_i(d_1, ..., d_m).
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1
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 19, 20, 29, 30, 39, 40, 42, 49, 50, 59, 60, 69, 70, 79, 80, 89, 90, 99, 100, 114, 115, 120, 121, 190, 199, 200, 207, 208, 210, 221, 260, 290, 299, 300, 301, 304, 330, 390, 399, 400, 420, 441, 448, 490, 499, 500, 572, 573, 590, 599, 600, 620
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OFFSET
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1,2
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COMMENTS
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Equivalently, integers k that are divisible by A061486(k); the quotients obtained when these terms k are divided by A061486(k) are in A343132.
The idea of this sequence comes from A328864 (1st problem of the 30th British Mathematical Olympiad in 1994) and also from the elementary symmetric polynomials in m variables (see links) that allow the generalization of this Olympiad problem to m-digit numbers.
-> When k = u has only one digit, then f_1(k) = e_1(u) = u and the numbers u from 1 to 9 satisfy u/e_1(u) = 1, so those are the first nine terms of this sequence.
-> When k = du has two digits, the numbers that are divisible by f_2(k) = (e_1(d,u) + e_2(d,u)) = (d+u) + (d*u) are the first 19 terms of A038366, from a(10) = 10 to a(28) = 99.
-> when k = hdu is a 3-digit number, then e_1(h,d,u) = h+d+u, e_2(h,d,u) = h*d+d*u+u*h and e_3(h,d,u) = h*d*u so f_3(k) = (h+d+u) + (hd+du+uh) + (hdu). This subsequence with 3-digit numbers A328864 has 50 terms that are here from a(29) = 100 to a(78) = 999.
-> When k = thdu is a 4-digit number, then e_1(t,h,d,u) = e_1(k) = t+h+d+u, e_2(t,h,d,u)= t*h+t*d+t*u+h*d+h*u+d*u, e_3(t,h,d,u) = t*h*d+t*h*u+t*d*u+h*d*u and e_4(t,h,d,u) = t*h*d*u with f_4(k) = e_1(t,h,d,u) + e_2(t,h,d,u) + e_3(t,h,d,u) + e_4(t,h,d,u). Numbers k such that k/f(k) is an integer form another subsequence with 87 terms. This subsequence with 4-digit numbers goes from a(79) = 1000 to a(165) = 9999.
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LINKS
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EXAMPLE
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For k = 7, f_1(7) = 7, and 7/7 = 1, hence 7 is a term.
For k = 42, f_2(4,2) = 4+2 + 4*2 = 14 and 42/14 = 3, hence 42 is a term.
For k = 572, f_3(5,7,2) = 5+7+2 + 5*7+7*2+2*5 + 5*7*2 = 14 + 59 + 70 = 143 and 572/143 = 4, hence 572 is a term.
For k = 3225, f_4(3,2,2,5) = 3+2+2+5 + 3*2+3*2+3*5+2*2+2*5+2*5 + 3*2*2+3*2*5+3*2*5+2*2*5 + 3*2*2*5 = 215 and 3225/215 = 15, hence 3225 is a term.
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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)); \\ A061486
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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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