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 A333979 Array read by antidiagonals, n >= 0, k >= 2: T(n,k) is the "digital derivative" of n in base k; if the base k representation of n is Sum_{j>=0} d_j*k^j, then T(n,k) = Sum_{j>=1} d_j*j*k^(j-1). 4
 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 4, 0, 0, 0, 0, 1, 4, 0, 0, 0, 0, 1, 1, 5, 0, 0, 0, 0, 0, 1, 2, 5, 0, 0, 0, 0, 0, 1, 1, 2, 12, 0, 0, 0, 0, 0, 0, 1, 1, 2, 12, 0, 0, 0, 0, 0, 0, 1, 1, 2, 6, 13, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 6, 13 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,15 LINKS Pontus von Brömssen, Antidiagonals n = 0..99, flattened M. J. Bannister, Z. Cheng, W. E. Devanny, and D. Eppstein, Superpatterns and universal point sets, Journal of Graph Algorithms and Applications 18(2) (2014), 177-209. FORMULA T(n,k) = floor(n/k) + k*T(floor(n/k),k). Proof: With n = Sum_{j>=0} d_j*k^j we have floor(n/k) + k*T(floor(n/k),k) = Sum_{j>=1} (d_j*k^(j-1) + k*d_j*(j-1)*k^(j-2)) = Sum_{j>=1} d_j*j*k^(j-1) = T(n,k). T(n,k) = T(n-1,k) + A055129(A286561(n,k),k). Proof: Let n = Sum_{j>=0} d_j*k^j and pick v so that d_j = 0 for j < v and d_v > 0 (so v = A286561(n,k)). Then n - 1 = sum_{j>=0} e_j*k^j, where e_j = k - 1 for j < v, e_v = d_v - 1, and e_j = d_j for j > v. We get T(n,k) - T(n-1,k) = Sum_{j>=1} j*(d_j-e_j)*k^(j-1) = v*k^(v-1) - (k-1)*Sum_{1<=j=0} k^j*floor(n/k**(j+1)). EXAMPLE Array begins:   n\k|  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16   ---|---------------------------------------------    0 |  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0    1 |  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0    2 |  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0    3 |  1  1  0  0  0  0  0  0  0  0  0  0  0  0  0    4 |  4  1  1  0  0  0  0  0  0  0  0  0  0  0  0    5 |  4  1  1  1  0  0  0  0  0  0  0  0  0  0  0    6 |  5  2  1  1  1  0  0  0  0  0  0  0  0  0  0    7 |  5  2  1  1  1  1  0  0  0  0  0  0  0  0  0    8 | 12  2  2  1  1  1  1  0  0  0  0  0  0  0  0    9 | 12  6  2  1  1  1  1  1  0  0  0  0  0  0  0   10 | 13  6  2  2  1  1  1  1  1  0  0  0  0  0  0   11 | 13  6  2  2  1  1  1  1  1  1  0  0  0  0  0   12 | 16  7  3  2  2  1  1  1  1  1  1  0  0  0  0   13 | 16  7  3  2  2  1  1  1  1  1  1  1  0  0  0   14 | 17  7  3  2  2  2  1  1  1  1  1  1  1  0  0   15 | 17  8  3  3  2  2  1  1  1  1  1  1  1  1  0   16 | 32  8  8  3  2  2  2  1  1  1  1  1  1  1  1 64 = 2*3^3 + 1*3^2 + 0*3^1 + 1*3^0, so T(64,3) = 2*3*3^2 + 1*2*3^1 + 0*1*3^0 = 60. Alternatively, using the formula T(n,k) = floor(n/k) + k*T(floor(n/k),k), we get T(64,3) = 21 + 3*T(21,3) = 21 + 3*(7 + 3*T(7,3)) = 42 + 9*(2 + 3*T(2,3)) = 60. PROG (Python) import sympy def A333979(n, k):   d=sympy.ntheory.factor_.digits(n, k)   return sum(j*d[-j-1]*k**(j-1) for j in range(1, len(d)-1)) (Python) # Second program (faster) def A333979(n, k):   return n//k+k*A333979(n//k, k) if n>=k else 0 CROSSREFS Cf. A136013 (column k=2), A080277 (every second term of column k=2), A080333 (every third term of column k=3). Cf. A007824, A055129, A286561. Sequence in context: A289110 A287287 A323519 * A276580 A331437 A351572 Adjacent sequences:  A333976 A333977 A333978 * A333980 A333981 A333982 KEYWORD nonn,base,tabl AUTHOR Pontus von Brömssen, Sep 04 2020 STATUS approved

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Last modified May 28 15:04 EDT 2022. Contains 354115 sequences. (Running on oeis4.)