A104258 -id:A104258 - cp's OEIS frontend

A341907 T(n, k) is the result of replacing 2^e with k^e in the binary expansion of n; square array T(n, k) read by antidiagonals upwards, n, k >= 0.

0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 2, 2, 1, 0, 1, 1, 3, 3, 1, 0, 0, 2, 4, 4, 4, 1, 0, 1, 2, 5, 9, 5, 5, 1, 0, 0, 3, 6, 10, 16, 6, 6, 1, 0, 1, 1, 7, 12, 17, 25, 7, 7, 1, 0, 0, 2, 8, 13, 20, 26, 36, 8, 8, 1, 0, 1, 2, 9, 27, 21, 30, 37, 49, 9, 9, 1, 0, 0, 3, 10, 28, 64, 31, 42, 50, 64, 10, 10, 1, 0

Offset: 0

Rémy Sigrist, Jun 04 2021

For any n >= 0, the n-th row, k -> T(n, k), corresponds to a polynomial in k with coefficients in {0, 1}.

For any k > 1, the k-th column, n -> T(n, k), corresponds to sums of distinct powers of k.

			Array T(n, k) begins:
  n\k|  0  1   2   3   4    5    6    7    8    9    10    11    12
  ---+-------------------------------------------------------------
    0|  0  0   0   0   0    0    0    0    0    0     0     0     0
    1|  1  1   1   1   1    1    1    1    1    1     1     1     1
    2|  0  1   2   3   4    5    6    7    8    9    10    11    12
    3|  1  2   3   4   5    6    7    8    9   10    11    12    13
    4|  0  1   4   9  16   25   36   49   64   81   100   121   144
    5|  1  2   5  10  17   26   37   50   65   82   101   122   145
    6|  0  2   6  12  20   30   42   56   72   90   110   132   156
    7|  1  3   7  13  21   31   43   57   73   91   111   133   157
    8|  0  1   8  27  64  125  216  343  512  729  1000  1331  1728
    9|  1  2   9  28  65  126  217  344  513  730  1001  1332  1729
   10|  0  2  10  30  68  130  222  350  520  738  1010  1342  1740
   11|  1  3  11  31  69  131  223  351  521  739  1011  1343  1741
   12|  0  2  12  36  80  150  252  392  576  810  1100  1452  1872

Rémy Sigrist, Table of n, a(n) for n = 0..10010
Index entries for sequences related to binary expansion of n

See A342707 for a similar sequence.

Cf. A000035, A000120, A000583, A000695, A001093, A002061, A002378, A002522, A002523, A005836, A007088, A011379, A027444, A033042, A033043, A033044, A033045, A033046, A033047, A033048, A033049, A034262, A053698, A071568, A098547, A104258.

T(n,k) = { my (v=0, e); while (n, n-=2^e=valuation(n,2); v+=k^e); v }

T(n, n) = A104258(n).
T(n, 0) = A000035(n).
T(n, 1) = A000120(n).
T(n, 2) = n.
T(n, 3) = A005836(n).
T(n, 4) = A000695(n).
T(n, 5) = A033042(n).
T(n, 6) = A033043(n).
T(n, 7) = A033044(n).
T(n, 8) = A033045(n).
T(n, 9) = A033046(n).
T(n, 10) = A007088(n).
T(n, 11) = A033047(n).
T(n, 12) = A033048(n).
T(n, 13) = A033049(n).
T(0, k) = 0.
T(1, k) = 1.
T(2, k) = k.
T(3, k) = k + 1.
T(4, k) = k^2.
T(5, k) = k^2 + 1 = A002522(k).
T(6, k) = k^2 + k = A002378(k).
T(7, k) = k^2 + k + 1 = A002061(k).
T(8, k) = k^3.
T(9, k) = k^3 + 1 = A001093(k).
T(10, k) = k^3 + k = A034262(k).
T(11, k) = k^3 + k + 1 = A071568(k).
T(12, k) = k^3 + k^2 = A011379(k).
T(13, k) = k^3 + k^2 + 1 = A098547(k).
T(14, k) = k^3 + k^2 + k = A027444(k).
T(15, k) = k^3 + k^2 + k + 1 = A053698(k).
T(16, k) = k^4 = A000583(k).
T(17, k) = k^4 + 1 = A002523(k).
T(m + n, k) = T(m, k) + T(n, k) when m AND n = 0 (where AND denotes the bitwise AND operator).

A193760 Replace 3^i with n^i in ternary representation of n.

Original entry on oeis.org

1, 2, 3, 5, 7, 12, 15, 18, 81, 101, 123, 156, 183, 212, 255, 289, 325, 648, 723, 802, 903, 991, 1083, 1200, 1301, 1406, 19683, 21953, 24391, 27030, 29823, 32802

Offset: 1

John Frye, Aug 04 2011

Cf. A104258 for the sequence formed by the binary rather than ternary numbers.

A193760 := proc(n) nb3 := convert(n,base,3) ; add(op(i,nb3)*n^(i-1),i=1..nops(nb3)) ; end proc: seq(A193760(n),n=1..40) ; # R. J. Mathar, Aug 05 2011

def A193760(n): return sum(d*n**i for i,d in enumerate(n.digits(base=3))) # D. S. McNeil, Aug 05 2011

A309801 If 2*n = Sum (2^e_k) then a(n) = Sum (e_k^n).

Original entry on oeis.org

1, 4, 9, 81, 244, 793, 2316, 65536, 262145, 1049600, 4196353, 17308657, 68703188, 273234809, 1088123500, 152587890625, 762939453126, 3814697527769, 19073486852414, 95370918425026, 476847618556329, 2384217176269538, 11921023106645561, 59886119752101281

Offset: 1

Ilya Gutkovskiy, Aug 17 2019

nonn

Replace 2^k with (k + 1)^n in binary representation of n.

			14 = 2*7 = 2^1 + 2^2 + 2^3 so a(7) = 1^7 + 2^7 + 3^7 = 2316.

Cf. A008935, A029931, A104258.

Table[Reverse[#].Range[Length[#]]^n &@IntegerDigits[n, 2], {n, 1, 24}]
Table[SeriesCoefficient[1/(1 - x) Sum[(k + 1)^n x^2^k/(1 + x^2^k), {k, 0, Floor[Log[2, n]] + 1}], {x, 0, n}], {n, 1, 24}]

a(n) = [x^n] (1/(1 - x)) * Sum_{k>=0} (k + 1)^n*x^(2^k)/(1 + x^(2^k)).

A356274 a(n) is the number whose base-(n+1) expansion equals the binary expansion of n.

Original entry on oeis.org

1, 3, 5, 25, 37, 56, 73, 729, 1001, 1342, 1741, 2366, 2941, 3615, 4369, 83521, 104977, 130340, 160021, 194922, 234741, 280393, 332377, 406250, 474553, 551151, 636637, 732511, 837901, 954304, 1082401, 39135393, 45435425, 52521910, 60466213, 69345326, 79236613

Offset: 1

Thomas Scheuerle, Aug 02 2022

If the binary expansion of n is n = bit0*2^0 + bit1*2^1 + bit2*2^2 + ... then a(n) = bit0*(n+1)^0 + bit1*(n+1)^1 + bit2*(n+1)^2 + ... . In other words: Interpret the binary expansion of n as digits in base n+1.

			n=4 in binary is 100 and interpreting those digits as base n+1 = 5 is a(4) = 25.

Cf. A000523, A007814, A104257, A104258, A128889, A136516.

a[n_] := FromDigits[IntegerDigits[n, 2], n + 1]; Array[a, 40] (* Amiram Eldar, Aug 19 2022 *)

PARI
```
a(n) = fromdigits(digits(n, 2), n+1)
```

def a(n): return sum((n+1)**i*int(bi) for i, bi in enumerate(bin(n)[2:][::-1]))
print([a(n) for n in range(1, 39)]) # Michael S. Branicky, Aug 02 2022

a(2^n) = (2^n + 1)^n = A136516(n).
a(2^n - 1) = (2^(n^2) - 1)/(2^n - 1) = A128889(n).
a(2^n + 1) = 1 + (2^n + 2)^n.
a(n) = A104257(n+1, n).
a(n) = (1/n)*Sum_{j>=1} floor((n + 2^(j-1))/2^j) * ((n-1)*(n+1)^(j-1) + 1).
a(n) = (1/n)*Sum_{j=1..n} ((n-1)*(n+1)^A007814(j) + 1).
a(2*n) = A104258(2*n+1) - 1.
(2*m+1)^n divides a((2*m+1)^n-1) for positive m and n > 0.

cp's OEIS Frontend

A341907 T(n, k) is the result of replacing 2^e with k^e in the binary expansion of n; square array T(n, k) read by antidiagonals upwards, n, k >= 0.

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A193760 Replace 3^i with n^i in ternary representation of n.

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A309801 If 2*n = Sum (2^e_k) then a(n) = Sum (e_k^n).

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A356274 a(n) is the number whose base-(n+1) expansion equals the binary expansion of n.

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