A048496
a(n) = 2^(n-1)*(3*n-4) + 3.
Original entry on oeis.org
1, 2, 7, 23, 67, 179, 451, 1091, 2563, 5891, 13315, 29699, 65539, 143363, 311299, 671747, 1441795, 3080195, 6553603, 13893635, 29360131, 61865987, 130023427, 272629763, 570425347, 1191182339, 2483027971, 5167382531
Offset: 0
n-th difference of a(n), a(n-1), ..., a(0) is (1, 4, 7, 10, ...).
A053565
a(n) = 2^(n-1)*(3*n-4).
Original entry on oeis.org
-2, -1, 4, 20, 64, 176, 448, 1088, 2560, 5888, 13312, 29696, 65536, 143360, 311296, 671744, 1441792, 3080192, 6553600, 13893632, 29360128, 61865984, 130023424, 272629760, 570425344, 1191182336, 2483027968, 5167382528, 10737418240
Offset: 0
- A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 189, 194-196.
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List([0..30], n-> 2^(n-1)*(3*n-4)) # G. C. Greubel, May 16 2019
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[2^(n-1)*(3*n-4): n in [0..30]]; // Vincenzo Librandi, Sep 26 2011
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Table[2^(n-1)*(3*n-4), {n,0,30}] (* G. C. Greubel, May 16 2019 *)
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vector(30, n, n--; 2^(n-1)*(3*n-4)) \\ G. C. Greubel, May 16 2019
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[2^(n-1)*(3*n-4) for n in (0..30)] # G. C. Greubel, May 16 2019
A104746
Array T(n,k) read by antidiagonals: T(1,k) = 2^k-1 and recursively T(n,k) = T(n-1,k) + A000337(k-1), n,k >= 1.
Original entry on oeis.org
1, 1, 3, 1, 4, 7, 1, 5, 12, 15, 1, 6, 17, 32, 31, 1, 7, 22, 49, 80, 63, 1, 8, 27, 66, 129, 192, 127, 1, 9, 32, 83, 178, 321, 448, 255, 1, 10, 37, 100, 227, 450, 769, 1024, 511, 1, 11, 42, 117, 276, 579, 1090, 1793, 2304, 1023, 1, 12, 47, 134, 325, 708, 1411, 2562, 4097, 5120, 2047, 1, 13, 52, 151, 374, 837, 1732, 3331, 5890, 9217, 11264, 4095
Offset: 1
To the first row, add the terms 0, 1, 5, 17, 49, 129, ... as indicated:
1, 3, 7, 15, 31, 63, ...
0, 1, 5, 17, 49, 129, ... (getting row 2 of the array:
1, 4, 12, 32, 80, 192, ... (= A001787, binomial transform for 1,2,3, ...)
Repeat the operation, getting the following array T(n,k):
1, 3, 7, 15, 31, 63, ...
1, 4, 12, 32, 80, 192, ...
1, 5, 17, 49, 129, 321, ...
1, 6, 22, 66, 178, 450, ...
-
A000337 := proc(n)
1+(n-1)*2^n ;
end proc:
A104746 := proc(n,k)
option remember;
if n= 1 then
2^k-1 ;
else
procname(n-1,k)+A000337(k-1) ;
end if;
end proc:
for d from 1 to 12 do
for k from 1 to d do
n := d-k+1 ;
printf("%d,",A104746(n,k)) ;
end do:
end do; # R. J. Mathar, Oct 30 2011
-
A000337[n_] := (n - 1)*2^n + 1;
T[1, k_] := 2^k - 1;
T[n_, k_] := T[n, k] = T[n - 1, k] + A000337[k - 1];
Table[T[n - k + 1, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 30 2024 *)
A271638
The total sum of the cubes of all parts of all compositions of n.
Original entry on oeis.org
1, 10, 48, 170, 512, 1398, 3580, 8770, 20808, 48206, 109652, 245850, 544864, 1196134, 2605164, 5636210, 12124280, 25952382, 55312516, 117440650, 248512656, 524288150, 1103102108, 2315255970, 4848615592, 10133438638, 21139292340, 44023414970, 91536490688
Offset: 1
The two compositions of n=2 are 2 and 1+1. The total sum of the cubes is a(2) = 2^3+1^3+1^3 = 10.
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Table[(13 n - 36) 2^(n - 1) + 6 n + 18, {n, 29}] (* or *)
Rest@ CoefficientList[Series[x (1 + 4 x + x^2)/((1 - 2 x) (1 - x))^2, {x, 0, 29}], x] (* Michael De Vlieger, Apr 11 2016 *)
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x='x+O('x^99); Vec(x*(1+4*x+x^2)/((2*x-1)*(1-x))^2) \\ Altug Alkan, Apr 11 2016
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for n in range(1,50):print((13*n-36)*2**(n-1)+6*n+18) # Soumil Mandal, Apr 11 2016
Showing 1-4 of 4 results.
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