A008854 -id:A008854 - cp's OEIS frontend

A191741 Dispersion of A047217, (numbers >1 and congruent to 0 or 1 or 2 mod 5), by antidiagonals.

1, 2, 3, 5, 6, 4, 10, 11, 7, 8, 17, 20, 12, 15, 9, 30, 35, 21, 26, 16, 13, 51, 60, 36, 45, 27, 22, 14, 86, 101, 61, 76, 46, 37, 25, 18, 145, 170, 102, 127, 77, 62, 42, 31, 19, 242, 285, 171, 212, 130, 105, 71, 52, 32, 23, 405, 476, 286, 355, 217, 176, 120

Offset: 1

Clark Kimberling, Jun 14 2011

For a background discussion of dispersions and their fractal sequences, see A191426.  For dispersions of congruence sequences mod 3, mod 4, or mod 5, see A191655, A191663, A191667, A191702.
...
Suppose that {2,3,4,5,6} is partitioned as {x1, x2} and {x3,x4,x5}.  Let S be the increasing sequence of numbers >1 and congruent to x1 or x2 mod 5, and let T be the increasing sequence of numbers >1 and congruent to x3 or x4 or x5 mod 5.  There are 10 sequences in S, each matched by a (nearly) complementary sequence in T.  Each of the 20 sequences generates a dispersion, as listed here:
...
A191722=dispersion of A008851 (0, 1 mod 5 and >1)
A191723=dispersion of A047215 (0, 2 mod 5 and >1)
A191724=dispersion of A047218 (0, 3 mod 5 and >1)
A191725=dispersion of A047208 (0, 4 mod 5 and >1)
A191726=dispersion of A047216 (1, 2 mod 5 and >1)
A191727=dispersion of A047219 (1, 3 mod 5 and >1)
A191728=dispersion of A047209 (1, 4 mod 5 and >1)
A191729=dispersion of A047221 (2, 3 mod 5 and >1)
A191730=dispersion of A047211 (2, 4 mod 5 and >1)
A191731=dispersion of A047204 (3, 4 mod 5 and >1)
...
A191732=dispersion of A047202 (2,3,4 mod 5 and >1)
A191733=dispersion of A047206 (1,3,4 mod 5 and >1)
A191734=dispersion of A032793 (1,2,4 mod 5 and >1)
A191735=dispersion of A047223 (1,2,3 mod 5 and >1)
A191736=dispersion of A047205 (0,3,4 mod 5 and >1)
A191737=dispersion of A047212 (0,2,4 mod 5 and >1)
A191738=dispersion of A047222 (0,2,3 mod 5 and >1)
A191739=dispersion of A008854 (0,1,4 mod 5 and >1)
A191740=dispersion of A047220 (0,1,3 mod 5 and >1)
A191741=dispersion of A047217 (0,1,2 mod 5 and >1)
...
For further information about these 20 dispersions, see A191722.
...
Regarding the dispersions A191722-A191741, there are general formulas for sequences of the type "(a or b mod m)" and "(a or b or c mod m)" used in the relevant Mathematica programs.

			Northwest corner:
1....2....5....10...17
3....6....11...20...35
4....7....12...21...36
8....15...26...45...76
9....16...27...46...77
13...22...37...62...105

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Cf. A047204, A047217, A191722, A191731, A191426.

(* Program generates the dispersion array t of the increasing sequence f[n] *)
r = 40; r1 = 12;  c = 40; c1 = 12;
a=2; b=5; c2=6; m[n_]:=If[Mod[n,3]==0,1,0];
f[n_]:=a*m[n+2]+b*m[n+1]+c2*m[n]+5*Floor[(n-1)/3]
Table[f[n], {n, 1, 30}]  (* A047217 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191741 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191741  *)

A194275 Concentric pentagonal numbers of the second kind: a(n) = floor(5n(n+1)/6).

Original entry on oeis.org

0, 1, 5, 10, 16, 25, 35, 46, 60, 75, 91, 110, 130, 151, 175, 200, 226, 255, 285, 316, 350, 385, 421, 460, 500, 541, 585, 630, 676, 725, 775, 826, 880, 935, 991, 1050, 1110, 1171, 1235, 1300, 1366, 1435, 1505, 1576, 1650, 1725, 1801, 1880, 1960, 2041, 2125

Offset: 0

Omar E. Pol, Aug 20 2011

Quasipolynomial: trisections are (15*x^2 - 15*x + 2)/2, 5*(15*x^2 - 5*x)/2, and 5*(15*x^2 + 5*x)/2. - Charles R Greathouse IV, Aug 23 2011
Appears to be similar to cellular automaton. The sequence gives the number of elements in the structure after n-th stage. Positive integers of A008854 gives the first differences. For a definition without words see the illustration of initial terms in the example section.
Also partial sums of A008854.
Also row sums of an infinite square array T(n,k) in which column k lists 3*k-1 zeros followed by the numbers A008706 (see example).
For concentric pentagonal numbers see A032527. - Omar E. Pol, Sep 27 2011

			Using the numbers A008706 we can write:
0, 1, 5, 10, 15, 20, 25, 30, 35, 40, 45, ...
0, 0, 0,  0,  1,  5, 10, 15, 20, 25, 30, ...
0, 0, 0,  0,  0,  0,  0,  1,  5, 10, 15, ...
0, 0, 0,  0,  0,  0,  0,  0,  0,  0,  1, ...
And so on.
===========================================
The sums of the columns give this sequence:
0, 1, 5, 10, 16, 25, 35, 46, 60, 75, 91, ...
...
Illustration of initial terms (in a precise representation the pentagons should appear strictly concentric):
.                                             o
.                                           o   o
.                            o            o       o
.                          o   o        o     o     o
.               o        o       o    o     o   o     o
.             o   o    o     o     o   o     o o     o
.      o    o       o   o         o     o           o
.    o   o   o     o     o       o       o         o
. o   o o     o o o       o o o o         o o o o o
.
. 1    5        10          16                25

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A000326, A008706, A008854, A032528, A152734, A193273, A193274.

Cf. similar sequences with the formula floor(k*n*(n+1)/(k+1)) listed in A281026.

[Floor(5*n*(n+1)/6): n in [0..60]]; // Vincenzo Librandi, Sep 27 2011

Table[Floor[5 n (n + 1)/6], {n, 0, 50}] (* Arkadiusz Wesolowski, Oct 03 2011 *)

a(n)=5*n*(n+1)\6 \\ Charles R Greathouse IV, Aug 23 2011

G.f.: (-1 - 3*x - x^2)/((-1 + x)^3*(1 + x + x^2)). - Alexander R. Povolotsky, Aug 22 2011

a(n) = floor(5*n*(n+1)/6). - Arkadiusz Wesolowski, Aug 23 2011

Name improved by Arkadiusz Wesolowski, Aug 23 2011

New name from Omar E. Pol, Sep 28 2011

A086458 Both n and n^3 have the same initial digit and also n and n^3 have the same final digit when expressed in base 10.

Original entry on oeis.org

0, 1, 10, 11, 29, 34, 99, 100, 101, 104, 105, 106, 109, 110, 111, 114, 115, 116, 119, 120, 121, 124, 125, 274, 275, 276, 279, 280, 281, 284, 285, 286, 289, 290, 291, 294, 295, 296, 299, 311, 314, 315, 316, 319, 320, 321, 324, 325, 326, 329, 330, 331, 334, 335

Offset: 0

Jeremy Gardiner, Jul 20 2003

Intersection of A008854 and A144582. - Michel Marcus, Mar 19 2015

			a(12) = 109 appears in the sequence because 109*109*109 = 1295029.

Harvey P. Dale, Table of n, a(n) for n = 0..2000

Cf. A086457 (similar sequence with squares).

Cf. A008854 (initial digit), A144582 (final digit).

sidQ[n_]:=Module[{idn=IntegerDigits[n],i3=IntegerDigits[n^3]},idn[[1]]==i3[[1]]&&idn[[-1]]== i3[[-1]]]; Select[Range[0,400],sidQ] (* Harvey P. Dale, May 14 2023 *)

isok(n) = (n == 0) || ((dn=digits(n)) && (ds=digits(n^3)) && (dn[#dn] == ds[#ds])); \\ Michel Marcus, Mar 19 2015

left$(str$(n), 1) = left$(str$(n^3), 1) AND right$(str$(n), 1) = right$(str$(n^3), 1)

A265187 Nonnegative m for which 2floor(m^2/11) = floor(2m^2/11).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 24, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 53, 54, 55, 56, 57, 59, 60, 61, 62, 64, 65, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 78, 79, 81, 82, 83, 84

Offset: 1

Bruno Berselli, Dec 04 2015

Also, nonnegative m not congruent to 3 or 8 (mod 11).
Integers x >= 0 satisfying k*floor(x^2/11) = floor(k*x^2/11) with k >= 0:
k = 0, 1:  x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...   (A001477);
k = 2:     x = 0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, ... (this sequence);
k = 3:     x = 0, 1, 5, 6, 10, 11, 12, 16, 17, 21, 22, ... (A265188);
k = 4..10: x = 0, 1, 10, 11, 12, 21, 22, 23, 32, 33, ...   (A112654);
k > 10:    x = 0, 11, 22, 33, 44, 55, 66, 77, 88, 99, ...  (A008593).
Primes in sequence: 2, 5, 7, 11, 13, 17, 23, 29, 31, 37, 43, 53, 59, ...

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,1,-1).

Cf. A008593, A112654, A265188.

Cf. similar sequences provided by 2*floor(m^2/h) = floor(2*m^2/h): A005843 (h=2), A001477 (h=3,4), A008854 (h=5), A047266 (h=6), A047299 (h=7), A042965 (h=8), A060464 (h=9), A237415 (h=10), this sequence (h=11), A047263 (h=12).

[n: n in [0..100] | 2*Floor(n^2/11) eq Floor(2*n^2/11)];

Select[Range[0, 100], 2 Floor[#^2/11] == Floor[2 #^2/11] &]
Select[Range[0, 100], ! MemberQ[{3, 8}, Mod[#, 11]] &]
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {0, 1, 2, 4, 5, 6, 7, 9, 10, 11}, 80]

is(n)=2*(n^2\11) == (2*n^2)\11 \\ Anders Hellström, Dec 05 2015

[n for n in (0..100) if 2*floor(n^2/11) == floor(2*n^2/11)]

G.f.: x^2*(1 + x + 2*x^2 + x^3 + x^4 + x^5 + 2*x^6 + x^7 + x^8)/((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)).

a(n) = a(n-1) + a(n-9) - a(n-10) for n>10.

A265188 Nonnegative m for which 3floor(m^2/11) = floor(3m^2/11).

Original entry on oeis.org

0, 1, 5, 6, 10, 11, 12, 16, 17, 21, 22, 23, 27, 28, 32, 33, 34, 38, 39, 43, 44, 45, 49, 50, 54, 55, 56, 60, 61, 65, 66, 67, 71, 72, 76, 77, 78, 82, 83, 87, 88, 89, 93, 94, 98, 99, 100, 104, 105, 109, 110, 111, 115, 116, 120, 121, 122, 126, 127, 131, 132, 133, 137, 138, 142

Offset: 1

Bruno Berselli, Dec 04 2015

See the second comment in A265187.
Also, nonnegative m congruent to 0, 1, 5, 6 or 10 (mod 11).
Primes in sequence: 5, 11, 17, 23, 43, 61, 67, 71, 83, 89, 109, 127, ...

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A265187.

Cf. similar sequences provided by 3*floor(n^2/h) = floor(3*n^2/h): A005843 (h=2), A008585 (h=3), A001477 (h=4), A008854 (h=5), A047266 (h=6), A047299 (h=7), A042965 (h=8), A265227 (h=9), A054967 (h=10), this sequence (h=11), A047266 (h=12).

[n: n in [0..150] | 3*Floor(n^2/11) eq Floor(3*n^2/11)];

Select[Range[0, 150], 3 Floor[#^2/11] == Floor[3 #^2/11] &]
Select[Range[0, 150], MemberQ[{0, 1, 5, 6, 10}, Mod[#, 11]] &]
LinearRecurrence[{1, 0, 0, 0, 1, -1}, {0, 1, 5, 6, 10, 11}, 70]

is(n) = 3*(n^2\11) == (3*n^2)\11 \\ Anders Hellström, Dec 05 2015

concat(0, Vec(x^2*(1 + 4*x + x^2 + 4*x^3 + x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^100))) \\ Michel Marcus, Dec 05 2015

[n for n in (0..150) if 3*floor(n^2/11) == floor(3*n^2/11)]

G.f.: x^2*(1 + 4*x + x^2 + 4*x^3 + x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).

a(n) = a(n-1) + a(n-5) - a(n-6), n>6.

A263536 Row sum of an equilateral triangle tiled with the 3,4,5 Pythagorean triple.

Original entry on oeis.org

5, 7, 12, 17, 19, 24, 29, 31, 36, 41, 43, 48, 53, 55, 60, 65, 67, 72, 77, 79, 84, 89, 91, 96, 101, 103, 108, 113, 115, 120, 125, 127, 132, 137, 139, 144, 149, 151, 156, 161, 163, 168, 173, 175, 180, 185, 187, 192, 197, 199, 204, 209, 211, 216, 221, 223, 228

Offset: 1

Craig Knecht, Oct 20 2015

Maximum number of Pythagorean triples in an equilateral triangle.
Two rules are used to construct this equilateral triangle: #1. Start with the number 5 at the top. #2. Require every "triple" to contain the Pythagorean triple 3, 4, 5 (see link below).
Up and down Pythagorean triples consist of two terms below and one above when k is odd (an up triple), and two terms above and one below when k is even (a down triple). Three adjacent terms in a straight line within the triangle form a linear triple.

			Triangle T(n,k):           Row sum
  5;                          5
  3, 4;                       7
  4, 5, 3;                   12
  5, 3, 4, 5;                17
  3, 4, 5, 3, 4;             19
  4, 5, 3, 4, 5, 3;          24

Colin Barker, Table of n, a(n) for n = 1..1000
Craig Knecht, Equilateral triangle tiled with 3,4,5 Pythagorean triples.
Craig Knecht, Interlocked up/down Pythagorean pairs.
Craig Knecht, Linear and triangular triples.
Craig Knecht, Incarcerated numbers.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A136289 (every triple contains 1,2,3), A008854 (every triple contains 1,2,2), A259052 (sum of Pascal triples).

Vec(x*(5*x^2+2*x+5)/((x-1)^2*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Oct 26 2015

From Colin Barker, Oct 26 2015: (Start)
a(n) = a(n-1)+a(n-3)-a(n-4) for n>4.
G.f.: x*(5*x^2+2*x+5) / ((x-1)^2*(x^2+x+1)).
(End)

A324561 Numbers with at least one prime index equal to 0, 1, or 4 modulo 5.

Original entry on oeis.org

2, 4, 6, 7, 8, 10, 11, 12, 13, 14, 16, 18, 20, 21, 22, 23, 24, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 38, 39, 40, 42, 43, 44, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 58, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 76, 77, 78, 80, 82, 84, 86, 87

Offset: 1

Gus Wiseman, Mar 06 2019

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.

Also Heinz numbers of the integer partitions counted by A039900. The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

			The sequence of terms together with their prime indices begins:
   2: {1}
   4: {1,1}
   6: {1,2}
   7: {4}
   8: {1,1,1}
  10: {1,3}
  11: {5}
  12: {1,1,2}
  13: {6}
  14: {1,4}
  16: {1,1,1,1}
  18: {1,2,2}
  20: {1,1,3}
  21: {2,4}
  22: {1,5}
  23: {9}
  24: {1,1,1,2}

Cf. A008854, A039900, A055396, A056239, A061395, A106529, A112798.

Cf. A324519, A324521, A324522, A324560, A324561, A324562.

with(numtheory):
q:= n-> is(irem(pi(min(factorset(n))), 5) in {0, 1, 4}):
select(q, [$2..100])[];  # Alois P. Heinz, Mar 07 2019

Select[Range[100],Intersection[Mod[If[#==1,{},PrimePi/@First/@FactorInteger[#]],5],{0,1,4}]!={}&]

A128788 a(n) = n^2*9^n.

Original entry on oeis.org

0, 9, 324, 6561, 104976, 1476225, 19131876, 234365481, 2754990144, 31381059609, 348678440100, 3797108212689, 40669853253264, 429575324987601, 4483851321172356, 46325504721296025, 474373168346071296

Offset: 0

Mohammad K. Azarian, Apr 07 2007

Subsequence of A008854 and A047462. [Bruno Berselli, Feb 07 2013]

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (27,-243,729).

Cf. A007758, A008854, A036289, A047462.

[n^2*9^n: n in [0..20]]; // Vincenzo Librandi, Feb 07 2013

CoefficientList[Series[9 x (1 + 9 x)/(1 - 9 x)^3, {x, 0, 30}], x] (* or *) LinearRecurrence[{27, -243, 729}, {0, 9, 324}, 20] (* Vincenzo Librandi, Feb 07 2013 *)

G.f.: 9*x*(1+9*x)/(1-9*x)^3. - Vincenzo Librandi, Feb 07 2013

a(n) = 27*a(n-1) - 243*a(n-2) + 729*a(n-3). - Vincenzo Librandi, Feb 07 2013

A299646 a(n) = Sum_{k = n..2*n+1} k^2.

Original entry on oeis.org

1, 14, 54, 135, 271, 476, 764, 1149, 1645, 2266, 3026, 3939, 5019, 6280, 7736, 9401, 11289, 13414, 15790, 18431, 21351, 24564, 28084, 31925, 36101, 40626, 45514, 50779, 56435, 62496, 68976, 75889, 83249, 91070, 99366, 108151, 117439, 127244, 137580, 148461, 159901

Offset: 0

Bruno Berselli, Feb 20 2018

Inverse binomial transform is 1, 13, 27, 14, 0, 0, 0, ... (0 continued).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Subsequence of A008854, A047388, A174070 (after 1).

Cf. A050409: Sum_{k = n..2*n} k^2; A050410: Sum_{k = n..2*n-1} k^2.

List([0..50], n -> (n+2)*(14*n^2+11*n+3)/6);

[(n+2)*(14*n^2+11*n+3)/6: n in [0..50]];

seq((n + 2)*(14*n^2 + 11*n + 3)/6, n=0..50); # Peter Luschny, Feb 21 2018

Table[(n + 2) (14 n^2 + 11 n + 3)/6, {n, 0, 50}]
(* Second program: *)
LinearRecurrence[{4, -6, 4, -1}, {1, 14, 54, 135}, 41] (* Jean-François Alcover, Feb 21 2018 *)

makelist((n+2)*(14*n^2+11*n+3)/6, n, 0, 50);

a(n)=(n+2)*(14*n^2+11*n+3)/6 \\ Charles R Greathouse IV, Feb 21 2018

Vec((1 + 10*x + 4*x^2 - x^3)/(1 - x)^4 + O(x^60)) \\ Colin Barker, Feb 22 2018

[(n+2)*(14*n^2+11*n+3)/6 for n in (0..50)]

O.g.f.: (1 + 10*x + 4*x^2 - x^3)/(1 - x)^4.
E.g.f.: (6 + 78*x + 81*x^2 + 14*x^3)*exp(x)/6.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (n + 2)*(14*n^2 + 11*n + 3)/6. Therefore:
a(6*k + r) = 504*k^3 + 18*(14*r + 13)*k^2 + (42*r^2 + 78*r + 25)*k + a(r), with 0 <= r <= 5. Example: for r=5, a(6*k + 5) = (6*k + 7)*(84*k^2 + 151*k + 68).

A374492 Nonsquares with last digit in {0, 1, 4, 5, 6, 9}.

Original entry on oeis.org

5, 6, 10, 11, 14, 15, 19, 20, 21, 24, 26, 29, 30, 31, 34, 35, 39, 40, 41, 44, 45, 46, 50, 51, 54, 55, 56, 59, 60, 61, 65, 66, 69, 70, 71, 74, 75, 76, 79, 80, 84, 85, 86, 89, 90, 91, 94, 95, 96, 99, 101, 104, 105, 106, 109, 110, 111, 114, 115, 116, 119, 120, 124, 125

Offset: 1

Stefano Spezia, Aug 04 2024

Squares have the last digit in {0, 1, 4, 5, 6, 9}, but the reverse is not always true since there are nonsquares that end with a digit in the same set. See Beiler.

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 139.

Robert Israel, Table of n, a(n) for n = 1..10000

Intersection of A000037 and A008854.

Cf. A000290.

remove(issqr, [seq(seq(10*a+b,b=[0,1,4,5,6,9]),a=0..20)]); # Robert Israel, Oct 30 2024

Select[Range[0,125],!IntegerQ[Sqrt[#]]&&MemberQ[{0,1,4,5,6,9},Last[IntegerDigits[#]]] &]

from math import isqrt
def ok(n): return n%10 in {0, 1, 4, 5, 6, 9} and isqrt(n)**2 < n
print([k for k in range(126) if ok(k)]) # Michael S. Branicky, Aug 04 2024

cp's OEIS Frontend

A191741 Dispersion of A047217, (numbers >1 and congruent to 0 or 1 or 2 mod 5), by antidiagonals.

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A194275 Concentric pentagonal numbers of the second kind: a(n) = floor(5*n*(n+1)/6).

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A086458 Both n and n^3 have the same initial digit and also n and n^3 have the same final digit when expressed in base 10.

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A265187 Nonnegative m for which 2*floor(m^2/11) = floor(2*m^2/11).

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A265188 Nonnegative m for which 3*floor(m^2/11) = floor(3*m^2/11).

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A263536 Row sum of an equilateral triangle tiled with the 3,4,5 Pythagorean triple.

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A324561 Numbers with at least one prime index equal to 0, 1, or 4 modulo 5.

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Comments

A194275 Concentric pentagonal numbers of the second kind: a(n) = floor(5n(n+1)/6).

A265187 Nonnegative m for which 2floor(m^2/11) = floor(2m^2/11).

A265188 Nonnegative m for which 3floor(m^2/11) = floor(3m^2/11).