A000302 -id:A000302 - cp's OEIS frontend

A301376 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that x^2-(3*y)^2 = 4^k for some k = 0,1,2,....

1, 1, 2, 1, 1, 3, 1, 1, 4, 2, 2, 3, 3, 3, 3, 1, 5, 6, 2, 2, 10, 5, 4, 3, 2, 7, 7, 3, 5, 4, 3, 1, 12, 8, 2, 6, 4, 5, 10, 2, 7, 13, 8, 5, 10, 6, 6, 3, 8, 4, 7, 7, 8, 11, 4, 3, 17, 9, 5, 4, 8, 5, 9, 1, 8, 14, 8, 8, 13, 5, 8, 6, 11, 10, 7, 5, 13, 15, 7, 2

Offset: 1

Zhi-Wei Sun, Mar 19 2018

nonn

Conjecture: a(n) > 0 for all n > 0. Moreover, any positive square n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers and y even such that x^2 - (3*y)^2 = 4^k for some k = 0,1,2,....
We have verifed this for all n = 1..10^7.
Compare this conjecture with the conjectures in A299537.
As 3*A001353(n)^2 + 1 = A001075(n)^2, the conjecture in A300441 implies that any positive square can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x^2 - 3*y^2 = 4^k for some k = 0,1,2,....
See also A301391 for a similar conjecture.

			a(1) = 1 since 1^2 = 1^2 + 0^2 + 0^2 + 0^2 with 1^2 - (3*0)^2 = 4^0.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4^2 - (3*0)^2 = 4^2.
a(7) = 1 since 7^2 = 2^2 + 0^2 + 3^2 + 6^2 with 2^2 - (3*0)^2 = 4^1.
a(31) = 3 since 31^2 = 10^2 + 2^2 + 4^2 + 29^2 with 10^2 - (3*2)^2 = 4^3, and 31^2 = 20^2 + 4^2 + 4^2 + 23^2 = 20^2 + 4^2 + 16^2 + 17^2 with 20^2 - (3*4)^2 = 4^4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000118, A000290, A000302, A299537, A299794, A299924, A300219, A300396, A300441, A300510, A301391.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[4^k+9y^2]&&QQ[n^2-4^k-10y^2],Do[If[SQ[n^2-(4^k+10y^2)-z^2],r=r+1],{z,0,Sqrt[(n^2-4^k-10y^2)/2]}]],{k,0,Log[2,n]},{y,0,Sqrt[(n^2-4^k)/10]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301391 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y even such that x^2 - (6*y)^2 = 4^k for some k = 0,1,2,....

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 2, 1, 3, 4, 1, 1, 8, 2, 2, 2, 3, 2, 6, 1, 2, 2, 1, 1, 11, 3, 2, 4, 4, 3, 3, 1, 6, 10, 6, 2, 7, 2, 3, 2, 6, 3, 8, 2, 7, 7, 2, 1, 11, 4, 4, 2, 2, 1, 6, 1, 3, 11, 3, 3, 16, 3, 5, 4, 8, 5, 2, 3, 11, 5, 8, 1

Offset: 1

Zhi-Wei Sun, Mar 20 2018

nonn

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 11, 13, 19, 2^k*m (k = 0,1,2,... and m = 1, 5, 7, 31).
We have verified a(n) > 0 for all n = 1..10^7.
See also A301376 for a similar conjecture.

			a(2) = 1 since 2^2 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 - (6*0)^2 = 4^1.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4^2 - (6*0)^2 = 4^2.
a(7) = 1 since 7^2 = 2^2 + 0^2 + 3^2 + 6^2 with 2^2 - (6*0)^2 = 4^1.
a(11) = 1 since 11 = 2^2 + 0^2 + 6^2 + 9^2 with 2^2 - (6*0)^2 = 4^1.
a(13) = 1 since 13 = 4^2 + 0^2 + 3^2 + 12^2 with 4^2 - (6*0)^2 = 4^2.
a(19) = 1 since 19 = 1^2 + 0^2 + 6^2 + 18^2 with 1^2 - (6*0)^2 = 4^0.
a(31) = 1 since 31^2 = 20^2 + 2^2 + 14^2 + 19^2 with 20^2 - (6*2)^2 = 4^4.
a(75) = 2 since 75^2 = 68^2 + 10^2 + 1^2 + 30^2 = 68^2 + 10^2 + 15^2 + 26^2 with 68^2 - (6*10)^2 = 4^5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000118, A000290, A000302, A299537, A299794, A299924, A300219, A300396, A300441, A300510, A301376.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[4^k+144m^2]&&QQ[n^2-4^k-148m^2], Do[If[SQ[n^2-(4^k+148m^2)-z^2],r=r+1],{z,0,Sqrt[(n^2-4^k-148m^2)/2]}]],{k,0,Log[2,n]},{m,0,Sqrt[(n^2-4^k)/148]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A345914 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum >= 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 19, 20, 21, 22, 24, 26, 27, 28, 30, 31, 32, 35, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 67, 69, 70, 72, 73, 74, 76, 79, 80, 82, 83, 84, 86, 87, 88

Offset: 1

Gus Wiseman, Jul 04 2021

nonn

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The sequence of terms together with the corresponding compositions begins:
     0: ()           19: (3,1,1)        40: (2,4)
     1: (1)          20: (2,3)          41: (2,3,1)
     2: (2)          21: (2,2,1)        42: (2,2,2)
     3: (1,1)        22: (2,1,2)        43: (2,2,1,1)
     4: (3)          24: (1,4)          44: (2,1,3)
     6: (1,2)        26: (1,2,2)        46: (2,1,1,2)
     7: (1,1,1)      27: (1,2,1,1)      47: (2,1,1,1,1)
     8: (4)          28: (1,1,3)        48: (1,5)
    10: (2,2)        30: (1,1,1,2)      50: (1,3,2)
    11: (2,1,1)      31: (1,1,1,1,1)    51: (1,3,1,1)
    12: (1,3)        32: (6)            52: (1,2,3)
    13: (1,2,1)      35: (4,1,1)        53: (1,2,2,1)
    14: (1,1,2)      36: (3,3)          54: (1,2,1,2)
    15: (1,1,1,1)    37: (3,2,1)        55: (1,2,1,1,1)
    16: (5)          38: (3,1,2)        56: (1,1,4)

The version for prime indices is A000027, counted by A000041.
These compositions are counted by A116406.
The case of non-Heinz numbers of partitions is A119899, counted by A344608.
The version for Heinz numbers of partitions is A344609, counted by A344607.
These are the positions of terms >= 0 in A344618.
The version for unreversed alternating sum is A345913.
The opposite (k <= 0) version is A345916.
The strict (k > 0) case is A345918.
The complement is A345920, counted by A294175.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A008549, A025047, A027187, A032443, A034871, A114121, A120452, A163493, A238279, A344650, A344743.

stc[n_]:=Differences[Prepend[Join@@Position[ Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Select[Range[0,100],sats[stc[#]]>=0&]

A345915 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum <= 0.

Original entry on oeis.org

0, 3, 6, 10, 12, 13, 15, 20, 24, 25, 27, 30, 36, 40, 41, 43, 46, 48, 49, 50, 51, 53, 54, 55, 58, 60, 61, 63, 72, 80, 81, 83, 86, 92, 96, 97, 98, 99, 101, 102, 103, 106, 108, 109, 111, 116, 120, 121, 123, 126, 136, 144, 145, 147, 150, 156, 160, 161, 162, 163

Offset: 1

Gus Wiseman, Jul 08 2021

nonn

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The sequence of terms together with the corresponding compositions begins:
     0: ()
     3: (1,1)
     6: (1,2)
    10: (2,2)
    12: (1,3)
    13: (1,2,1)
    15: (1,1,1,1)
    20: (2,3)
    24: (1,4)
    25: (1,3,1)
    27: (1,2,1,1)
    30: (1,1,1,2)
    36: (3,3)
    40: (2,4)
    41: (2,3,1)

The version for Heinz numbers of partitions is A028260 (counted by A027187).
These compositions are counted by A058622.
These are the positions of terms <= 0 in A124754.
The reverse-alternating version is A345916.
The opposite (k >= 0) version is A345917.
The strictly negative (k < 0) version is A345919.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000097, A000346, A008549, A025047, A032443, A114121, A163493, A344607, A344609, A344610.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[stc[#]]<=0&]

A345916 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum <= 0.

Original entry on oeis.org

0, 3, 5, 9, 10, 13, 15, 17, 18, 23, 25, 29, 33, 34, 36, 39, 41, 43, 45, 46, 49, 50, 53, 55, 57, 58, 61, 63, 65, 66, 68, 71, 75, 77, 78, 81, 85, 89, 90, 95, 97, 98, 103, 105, 109, 113, 114, 119, 121, 125, 129, 130, 132, 135, 136, 139, 141, 142, 145, 147, 149

Offset: 1

Gus Wiseman, Jul 08 2021

nonn

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The sequence of terms together with the corresponding compositions begins:
     0: ()
     3: (1,1)
     5: (2,1)
     9: (3,1)
    10: (2,2)
    13: (1,2,1)
    15: (1,1,1,1)
    17: (4,1)
    18: (3,2)
    23: (2,1,1,1)
    25: (1,3,1)
    29: (1,1,2,1)
    33: (5,1)
    34: (4,2)
    36: (3,3)

The version for Heinz numbers of partitions is A000290.
These compositions are counted by A058622.
These are the positions of terms <= 0 in A344618.
The opposite (k >= 0) version is A345914.
The version for unreversed alternating sum is A345915.
The strictly negative (k < 0) version is A345920.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A008549, A025047, A027187, A028260, A032443, A114121, A163493, A344607, A344610, A345908.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Select[Range[0,100],sats[stc[#]]<=0&]

A003992 Square array read by upwards antidiagonals: T(n,k) = n^k for n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 9, 8, 1, 0, 1, 5, 16, 27, 16, 1, 0, 1, 6, 25, 64, 81, 32, 1, 0, 1, 7, 36, 125, 256, 243, 64, 1, 0, 1, 8, 49, 216, 625, 1024, 729, 128, 1, 0, 1, 9, 64, 343, 1296, 3125, 4096, 2187, 256, 1, 0, 1, 10, 81, 512, 2401, 7776, 15625, 16384, 6561, 512, 1, 0

Offset: 0

Marc LeBrun

If the array is transposed, T(n,k) is the number of oriented rows of n colors using up to k different colors. The formula would be T(n,k) = [n==0] + [n>0]*k^n. The generating function for column k would be 1/(1-k*x). For T(3,2)=8, the rows are AAA, AAB, ABA, ABB, BAA, BAB, BBA, and BBB. - Robert A. Russell, Nov 08 2018

T(n,k) is the number of multichains of length n from {} to [k] in the Boolean lattice B_k. - Geoffrey Critzer, Apr 03 2020

			Rows begin:
[1, 0,  0,   0,    0,     0,      0,      0, ...],
[1, 1,  1,   1,    1,     1,      1,      1, ...],
[1, 2,  4,   8,   16,    32,     64,    128, ...],
[1, 3,  9,  27,   81,   243,    729,   2187, ...],
[1, 4, 16,  64,  256,  1024,   4096,  16384, ...],
[1, 5, 25, 125,  625,  3125,  15625,  78125, ...],
[1, 6, 36, 216, 1296,  7776,  46656, 279936, ...],
[1, 7, 49, 343, 2401, 16807, 117649, 823543, ...], ...

T. D. Noe, Rows n = 0..50 of triangle, flattened

Rows 0-49 are A000007, A000012, A000079, A000244, A000302, A000351, A000400, A000420, A001018, A001019, A011557, A001020, A001021, A001022, A001023, A001024, A001025, A001026, A001027, A001029, A009964-A009992, A087752.
Columns 0-26 are A000012, A001477, A000290, A000578, A000583, A000584, A001014, A001015, A001016, A001017, A008454, A008455, A008456, A010801-A010813, A089081.
Main diagonal is A000312. Other diagonals include A000169, A007778, A000272, A008788. Antidiagonal sums are in A026898.
Cf. A099555.
Transpose is A004248. See A051128, A095884, A009999 for other versions.
Cf. A277504 (unoriented), A293500 (chiral).

[[(n-k)^k: k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 08 2018

Table[If[k == 0, 1, (n - k)^k], {n, 0, 11}, {k, 0, n}]//Flatten

T(n,k) = (n-k)^k \\ Charles R Greathouse IV, Feb 07 2017

E.g.f.: Sum T(n,k)*x^n*y^k/k! = 1/(1-x*exp(y)). - Paul D. Hanna, Oct 22 2004

E.g.f.: Sum T(n,k)*x^n/n!*y^k/k! = e^(x*e^y). - Franklin T. Adams-Watters, Jun 23 2006

More terms from David W. Wilson

Edited by Paul D. Hanna, Oct 22 2004

A034870 Even-numbered rows of Pascal's triangle.

Original entry on oeis.org

1, 1, 2, 1, 1, 4, 6, 4, 1, 1, 6, 15, 20, 15, 6, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1

Offset: 0

N. J. A. Sloane

The sequence of row lengths of this array is [1,3,5,7,9,11,13,...]= A005408(n), n>=0.
Equals X^n * [1,0,0,0,...] where X = an infinite tridiagonal matrix with (1,1,1,...) in the main and subsubdiagonal and (2,2,2,...) in the main diagonal. X also = a triangular matrix with (1,2,1,0,0,0,...) in each column. - Gary W. Adamson, May 26 2008
a(n,m) has the following interesting combinatoric interpretation. Let s(n,m) equal the set of all base-4, n-digit numbers with n-m more 1-digits than 2-digits. For example s(2,1) = {10,01,13,31} (note that numbers like 1 are left-padded with 0's to ensure that they have 2 digits). Notice that #s(2,1) = a(2,1) with # indicating cardinality. This is true in general. a(n,m)=#s(n,m). In words, a(n,m) gives the number of n-digit, base-4 numbers with n-m more 1 digits than 2 digits. A proof is provided in the Links section. - Russell Jay Hendel, Jun 23 2015

			Triangle begins:
  1;
  1,  2,  1;
  1,  4,  6,   4,   1;
  1,  6, 15,  20,  15,   6,   1;
  1,  8, 28,  56,  70,  56,  28,   8,   1;
  1, 10, 45, 120, 210, 252, 210, 120,  45,  10,  1;
  1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1;

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Peter Bala, Notes on generalized Riordan arrays
E. H. M. Brietzke, An identity of Andrews and a new method for the Riordan array proof of combinatorial identities, Discrete Math., 308 (2008), 4246-4262.
Russell Jay Hendel, Proof that a(n,m) gives the number of n-digit, base-4 numbers with n-m more 1-digits than 2-digits.
Wolfdieter Lang, First 9 rows.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A bracket polynomial for 2-tangle shadows, arXiv:2002.06672 [math.CO], 2020.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A034871.
Cf. A000302 (row sums, powers of 4), alternating row sums are 0, except for n=0 which gives 1.
Cf. A003095, A034839, A080928, A086645.

a034870 n k = a034870_tabf !! n !! k
a034870_row n = a034870_tabf !! n
a034870_tabf = map a007318_row [0, 2 ..]
-- Reinhard Zumkeller, Apr 19 2012, Apr 02 2011

/* As triangle: */ [[Binomial(n,k): k in [0..n]]: n in [0.. 15 by 2]]; // Vincenzo Librandi, Jul 16 2015

T := (n,k) -> simplify(GegenbauerC(`if`(kPeter Luschny, May 08 2016

Flatten[Table[Binomial[n,k],{n,0,20,2},{k,0,n}]] (* Harvey P. Dale, Dec 15 2014 *)

taylor(1/(1-x*(y+1)^2),x,0,10,y,0,10); /* Vladimir Kruchinin, Nov 22 2020 */

flatten([[binomial(2*n, k) for k in (0..2*n)] for n in (0..12)]) # G. C. Greubel, Mar 18 2022

T(n, m) = binomial(2*n, m), 0<= m <= 2*n, 0<=n, else 0.
G.f. for column m=2*k sequence: (x^k)*Pe(k, x)/(1-x)^(2*k+1), k>=0; for column m=2*k-1 sequence (x^k)*Po(k, x)/(1-x)^(2*k), k>=1, with the row polynomials Pe(k, x) := sum(A091042(k, m)*x^m, m=0..k) and Po(k, x) := 2*sum(A091044(k, m)*x^m, m=0..k-1); see also triangle A091043.
From Paul D. Hanna, Apr 18 2012: (Start)
Let A(x) be the g.f. of the flattened sequence, then:
G.f.: A(x) = Sum_{n>=0} x^(n^2) * (1+x)^(2*n).
G.f.: A(x) = Sum_{n>=0} x^n*(1+x)^(2*n) * Product_{k=1..n} (1 - (1+x)^2*x^(4*k-3)) / (1 - (1+x)^2*x^(4*k-1)).
G.f.: A(x) = 1/(1 - x*(1+x)^2/(1 + x*(1-x^2)*(1+x)^2/(1 - x^5*(1+x)^2/(1 + x^3*(1-x^4)*(1+x)^2/(1 - x^9*(1+x)^2/(1 + x^5*(1-x^6)*(1+x)^2/(1 - x^13*(1+x)^2/(1 + x^7*(1-x^8)*(1+x)^2/(1 - ...))))))))), a continued fraction.
(End)
From Peter Bala, Jul 14 2015: (Start)
Denote this array by P. Then P * transpose(P) is the square array ( binomial(2*n + 2*k, 2*k) )n,k>=0, which, read by antidiagonals, is A086645.
Transpose(P) is a generalized Riordan array (1, (1 + x)^2) as defined in the Bala link.
Let p(x) = (1 + x)^2. P^2 gives the coefficients in the expansion of the polynomials ( p(p(x)) )^n, P^3 gives the coefficients in the expansion of the polynomials ( p(p(p(x))) )^n and so on.
Row sums are 2^(2*n); row sums of P^2 are 5^(2*n), row sums of P^3 are 26^(2*n). In general, the row sums of P^k, k = 0,1,2,..., are equal to A003095(k)^(2*n).
The signed version of this array ( (-1)^k*binomial(2*n,k) )n,k>=0 is a left-inverse for A034839.
A034839 * P = A080928. (End)
T(n, k) = GegenbauerC(m, -n, -1) where m = k if kPeter Luschny, May 08 2016
G.f.: 1/(1-x*(y+1)^2). - Vladimir Kruchinin, Nov 22 2020

A163355 Permutation of integers for constructing Hilbert curve in N x N grid.

Original entry on oeis.org

0, 1, 3, 2, 14, 15, 13, 12, 4, 7, 5, 6, 8, 11, 9, 10, 16, 19, 17, 18, 20, 21, 23, 22, 30, 29, 31, 28, 24, 25, 27, 26, 58, 57, 59, 56, 54, 53, 55, 52, 60, 61, 63, 62, 50, 51, 49, 48, 32, 35, 33, 34, 36, 37, 39, 38, 46, 45, 47, 44, 40, 41, 43, 42, 234, 235, 233, 232, 236, 239

Offset: 0

Antti Karttunen, Jul 29 2009

nonn

Antti Karttunen, Table of n, a(n) for n = 0..262143
Index entries for sequences that are permutations of the natural numbers

Inverse: A163356. A163357 & A163359 give two variants of Hilbert curve in N x N grid. Cf. also A163332.
Second and third "powers": A163905, A163915.
In range [A000302(n-1)..A024036(n)] of this permutation, the number of cycles is given by A163910, number of fixed points seems to be given by A147600(n-1) (fixed points themselves: A163901). Max. cycle sizes is given by A163911 and LCM's of all cycle sizes by A163912.
See also: A000695, A163890, A163894, A163902-A163903, A163914, A163485, A302843, A302845.

A057300 := proc(n)
    option remember;
    `if`(n=0, 0, procname(iquo(n, 4, 'r'))*4+[0, 2, 1, 3][r+1])
end proc:
A163355 := proc(n)
    option remember ;
    local d,base4,i,r ;
    if n <= 1 then
        return n ;
    end if;
    base4 := convert(n,base,4) ;
    d := op(-1,base4) ;
    i := nops(base4)-1 ;
    r := n-d*4^i ;
    if ( d=1 and type(i,even) ) or ( d=2 and type(i,odd)) then
        4^i+procname(A057300(r)) ;
    elif d= 3 then
        2*4^i+procname(A057300(r)) ;
    else
        3*4^i+procname(4^i-1-r) ;
    end if;
end proc:
seq(A163355(n),n=0..100) ; # R. J. Mathar, Nov 22 2023

A057300(n) = { my(t=1, s=0); while(n>0,  if(1==(n%4),n++,if(2==(n%4),n--)); s += (n%4)*t; n >>= 2; t <<= 2); (s); };
A163355(n) = if(!n,n,my(i = (#binary(n)-1)\2, f = 4^i, d = (n\f)%4, r = (n%f)); if(((1==d)&&!(i%2))||((2==d)&&(i%2)), f+A163355(A057300(r)), if(3==d,f+f+A163355(A057300(r)), (3*f)+A163355(f-1-r)))); \\ Antti Karttunen, Apr 14 2018

a(0) = 0, and given d=1, 2 or 3, then a((d*(4^i))+r)
  = (4^i) + a(A057300(r)), if d=1 and i is even, or if d=2 and i is odd
  = 2*(4^i) + a(A057300(r)), if d=3,
  = 3*(4^i) + a((4^i)-1-r) in other cases.
From Alan Michael Gómez Calderón, May 06 2025: (Start)
a(3*A000695(n)) = 2*A000695(n);
a(3*(A000695(n) + 2^A000695(2*m))) = 2*(A000695(n) + 2^A000695(2*m)) for m >= 2;
a((2 + 16^n)*2^(-1 + 4*m)) = 4^(2*(n + m) - 1) + (11*16^m - 2)/3. (End)

Links to further derived sequences added by Antti Karttunen, Sep 21 2009

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A346697 Sum of the odd-indexed parts (odd bisection) of the multiset of prime indices of n.

Original entry on oeis.org

0, 1, 2, 1, 3, 1, 4, 2, 2, 1, 5, 3, 6, 1, 2, 2, 7, 3, 8, 4, 2, 1, 9, 2, 3, 1, 4, 5, 10, 4, 11, 3, 2, 1, 3, 3, 12, 1, 2, 2, 13, 5, 14, 6, 5, 1, 15, 4, 4, 4, 2, 7, 16, 3, 3, 2, 2, 1, 17, 3, 18, 1, 6, 3, 3, 6, 19, 8, 2, 5, 20, 4, 21, 1, 5, 9, 4, 7, 22, 5, 4, 1

Offset: 1

Gus Wiseman, Aug 01 2021

nonn

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

			The prime indices of 1100 are {1,1,3,3,5}, so a(1100) = 1 + 3 + 5 = 9.
The prime indices of 2100 are {1,1,2,3,3,4}, so a(2100) = 1 + 2 + 3 = 6.

The version for standard compositions is A209281(n+1) (even: A346633).
Subtracting the even version gives A316524 (reverse: A344616).
The even version is A346698.
The reverse version is A346699.
The even reverse version is A346700.
A000120 and A080791 count binary digits 1 and 0, with difference A145037.
A000302 counts compositions with odd alternating sum, ranked by A053738.
A001414 adds up prime factors, row sums of A027746.
A029837 adds up parts of standard compositions (alternating: A124754).
A056239 adds up prime indices, row sums of A112798.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A325534 counts separable partitions, ranked by A335433.
A325535 counts inseparable partitions, ranked by A335448.
A344606 counts alternating permutations of prime indices.
Cf. A000070, A025047, A120452, A341446, A344614, A344617, A344653, A344654, A345957, A345958, A345959.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Table[Total[First/@Partition[Append[primeMS[n],0],2]],{n,100}]

a(n) = A056239(n) - A346698(n).
a(n) = A316524(n) + A346698(n).
a(n odd omega) = A346699(n).
a(n even omega) = A346700(n).
A344616(n) = A346699(n) - A346700(n).

cp's OEIS Frontend

A301376 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that x^2-(3*y)^2 = 4^k for some k = 0,1,2,....

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A301391 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y even such that x^2 - (6*y)^2 = 4^k for some k = 0,1,2,....

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A345914 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum >= 0.

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A345915 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum <= 0.

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A345916 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum <= 0.

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A003992 Square array read by upwards antidiagonals: T(n,k) = n^k for n >= 0, k >= 0.

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PARI

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A034870 Even-numbered rows of Pascal's triangle.

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A163355 Permutation of integers for constructing Hilbert curve in N x N grid.

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