A022554 -id:A022554 - cp's OEIS frontend

A363499 a(n) = Sum_{k=0..n} floor(sqrt(k))^5.

0, 1, 2, 3, 35, 67, 99, 131, 163, 406, 649, 892, 1135, 1378, 1621, 1864, 2888, 3912, 4936, 5960, 6984, 8008, 9032, 10056, 11080, 14205, 17330, 20455, 23580, 26705, 29830, 32955, 36080, 39205, 42330, 45455, 53231, 61007, 68783, 76559, 84335, 92111, 99887

Offset: 0

Hans J. H. Tuenter, Jun 05 2023

Partial sums of the fifth powers of the terms of A000196.

Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000

Sums of powers of A000196: A022554 (1st), A174060 (2nd), A363497 (3rd), A363498 (4th), this sequence (5th).

Table[(n + 1) #^5 - (1/84) # (# + 1)*(2 # + 1)*(3 # - 1)*(10 #^3 - 7 # + 4) &[Floor@ Sqrt[n]], {n, 0, 42}] (* Michael De Vlieger, Jun 10 2023 *)

from math import isqrt
def A363499(n): return (m:=isqrt(n))**5 *(n+1) - (m*(m+1)*(2*m+1)*(3*m-1)*(10*m**3-7*m+4))//84 # Karl-Heinz Hofmann, Jul 17 2023

a(n) = (n+1)*m^5 - (1/84)*m*(m+1)*(2*m+1)*(3*m-1)*(10*m^3-7*m+4), where m = floor(sqrt(n)).

A176615 Number of edges in the graph on n vertices, labeled 1 to n, where two vertices are joined just if their labels sum to a perfect square.

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 16, 17, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 49, 52, 55, 57, 59, 61, 63, 65, 68, 71, 74, 77, 80, 83, 86, 89, 91, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 127, 131, 135, 138, 141, 144, 147, 150

Offset: 1

Franklin T. Adams-Watters, Apr 21 2010

Equivalently, number of pairs of integers 0 < i < j <= n such that i + j is a square.

Suggested by R. K. Guy

			For n = 7 the graph contains the 4 edges 1-3, 2-7, 3-6, 4-5.

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Cf. A000196, A022554, A103128, A281706.

Column k=2 of A281871.

b:= n-> 1+floor(sqrt(2*n-1))-ceil(sqrt(n+1)):
a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+b(n)) end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 30 2017

a[n_] := Sum[Floor[Sqrt[2k-1]] - Floor[Sqrt[k]], {k, 1, n}]; Table[a[n], {n, 1, 68}] (* Jean-François Alcover, Nov 04 2011, after Pari *)

a(n)=sum(k=1,sqrtint(n+1),ceil(k^2/2)-1)+sum(k=sqrtint(n+1)+1,sqrtint(2*n -1),n-floor(k^2/2))

a(n)=sum(k=1,n,sqrtint(2*k-1)-sqrtint(k))

Asymptotically, a(n) ~ (2*sqrt(2) - 2)/3 n^(3/2). The error term is probably O(n^(1/2)); O(n) is easily provable.

A214036 Numbers k such that floor(sqrt(1)) + floor(sqrt(2)) + floor(sqrt(3)) + ... + floor(sqrt(k)) is prime.

Original entry on oeis.org

2, 3, 4, 5, 7, 8, 10, 14, 36, 37, 39, 42, 43, 44, 46, 47

Offset: 1

Paolo P. Lava, Mar 06 2013

The sequence is complete. Indeed, let s(n) be the sum of floor(sqrt(k)) for k from 1 to n. It is easy to verify that s(n^2+j), for 0 <= j < (n+1)^2-n^2, is equal to n(j+1) + n(4n+1)(n-1)/6, which is always divisible by n or by n/6 for n > 6. - Giovanni Resta, Mar 26 2014

			2 is a term because floor(sqrt(1))+floor(sqrt(2)) = 1+1 = 2 is prime;
14 is a term because floor(sqrt(1))+ ... +floor(sqrt(14)) = 1+1+1+2+2+2+2+2+3+3+3+3+3+3 = 31 is prime.

Primes in A022554.

Cf. A220953.

A214036:=proc(q)  local a,n; a:=0;
for n from 1 to q do a:=a+floor(sqrt(n)); if isprime(a) then print(n); fi; od; end:
A214036(10^10);
Alternative program:
A214036_bis:=proc(q)  local a,j,n; a:=0;
for n from 1 to q do for j from 1 to 2*n+1 do
    a:=a+n; if isprime(a) then print(n^2+j-1); fi;
od; od; end:
A214036_bis(10^10);

Position[Accumulate[Table[Floor[Sqrt[n]],{n,50}]],?PrimeQ]//Flatten (* _Harvey P. Dale, Apr 14 2017 *)

sm = 0; for (n=1, 10^9, sm+=sqrtint(n); if (isprime(sm), print1(n,", ")));
/* Joerg Arndt, Mar 07 2013 */

A270370 a(n) = Sum_{k=0..n} (-1)^k*floor(k^(1/3)).

Original entry on oeis.org

0, -1, 0, -1, 0, -1, 0, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 2, -2, 2, -2, 2, -2

Offset: 0

John M. Campbell, Mar 15 2016

			a(5) = [0^(1/3)]-[1^(1/3)]+[2^(1/3)]-[3^(1/3)]+[4^(1/3)]-[5^(1/3)] = 0-1+1-1+1-1 = -1, letting [] denote the floor function.

Cf. A268173, A022554, A031876, A032512, A032513, A032514, A032515, A032516, A032517, A032518, A032519, A032520, A032521.

Print[Table[Sum[(-1)^i*Floor[i^(1/3)],{i,0,n}],{n,0,100}]]

PARI
```
a(n)=sum(i=0,n,(-1)^i*sqrtnint(i,3))
```

a(n)=sqrtnint(n,3)*(-1)^n/2-((-1)^(sqrtnint(n,3)+1)+1)/4

a(n) = floor(n^(1/3))*(-1)^n/2 - ((-1)^(floor(n^(1/3))+1)+1)/4.

A327672 a(n) = Sum_{k=0..n} ceiling(sqrt(k)).

Original entry on oeis.org

0, 1, 3, 5, 7, 10, 13, 16, 19, 22, 26, 30, 34, 38, 42, 46, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 101, 107, 113, 119, 125, 131, 137, 143, 149, 155, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 260, 268, 276, 284, 292, 300, 308, 316

Offset: 0

Peter Kagey, Sep 21 2019

nonn

Partial sums of A003059.

Given a digraph whose vertices are numbered from 0 to n and in which an edge (u,v) exists iff u < v, a(n) is the maximum number of arcs that can be chosen so that for each vertex j other than 0 and n, the number of chosen arcs whose tail is vertex j equals the number of chosen arcs whose head is vertex j. - Xutong Ding, Dec 12 2023

Peter Kagey, Table of n, a(n) for n = 0..10000

Cf. A003059, A022554.

Accumulate[Ceiling[Sqrt[Range[0, 60]]]]
Table[(1 + Floor[Sqrt[n]])*(6*n - Floor[Sqrt[n]] - 2*Floor[Sqrt[n]]^2)/6, {n, 0, 100}] (* Vaclav Kotesovec, Dec 26 2023 *)

a(n) = (1 + floor(sqrt(n)))*(6*n - floor(sqrt(n)) - 2*floor(sqrt(n))^2)/6. - Vaclav Kotesovec, Dec 26 2023

A262352 a(n) = Sum_{k=0..n} (-1)^k*floor(k^(1/4)).

Original entry on oeis.org

0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1, -2, 1

Offset: 0

John M. Campbell, Mar 24 2016

			Letting [] denote the floor function, a(7) = [0^(1/4)] - [1^(1/4)] + [2^(1/4)] - [3^(1/4)] + [4^(1/4)] - [5^(1/4)] + [6^(1/4)] - [7^(1/4)] = 0 - 1 + 1 - 1 + 1 - 1 + 1 - 1 = -1.

Antti Karttunen, Table of n, a(n) for n = 0..16383
Antti Karttunen, Data supplement: n, a(n) computed for n = 0..65537

Cf. A270370, A268173, A022554, A031876, A032512, A032513, A032514, A032515, A032516, A032517, A032518, A032519, A032520, A032521.

Print[Table[Sum[(-1)^k*Floor[k^(1/4)],{k,0,n}],{n,0,100}]] ;

a(n)=floor(n^(1/4))*(-1)^n/2-((-1)^(floor(n^(1/4))+1)+1)/4

PARI
```
a(n)=sum(k=0,n,(-1)^k*floor(k^(1/4)))
```

A262352(n) = sum(k=0,n,((-1)^k)*sqrtnint(k, 4)); \\ Antti Karttunen, Nov 06 2018

a(n) = floor(n^(1/4))*(-1)^n/2-((-1)^(floor(n^(1/4))+1)+1)/4.

More terms from Antti Karttunen, Nov 06 2018

A331478 Irregular triangle T(n,k) = n - (s - k + 1)^2 for 1 <= k <= s, with s = floor(sqrt(n)).

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 2, 5, 3, 6, 4, 7, 0, 5, 8, 1, 6, 9, 2, 7, 10, 3, 8, 11, 4, 9, 12, 5, 10, 13, 6, 11, 14, 0, 7, 12, 15, 1, 8, 13, 16, 2, 9, 14, 17, 3, 10, 15, 18, 4, 11, 16, 19, 5, 12, 17, 20, 6, 13, 18, 21, 7, 14, 19, 22, 8, 15, 20, 23, 0, 9, 16, 21, 24, 1

Offset: 1

Michael De Vlieger, Jan 17 2020

Row n begins with n - floor(sqrt(n)).
Zero appears in row n for n that are perfect squares. Let r = sqrt(n). For perfect square n, there exists a partition of n that consists of a run of r parts that are each r themselves; e.g., for n = 4, we have {2, 2}, for n = 9, we have {3, 3, 3}. It is clear through the Ferrers diagram of these partitions that they are equivalent to their Durfee square, thus n - s^2 = 0.
Since the partitions of any n contain Durfee squares in the range of 1 <= s <= floor(sqrt(n)) (with perfect square n also including k = 0), the distinct Durfee square excesses must be the differences n - s^2 for 1 <= s <= floor(sqrt(n)).
We borrow the term "square excess" from A053186(n), which is simply the difference n - floor(sqrt(n)).
Row n of this sequence contains distinct Durfee square excesses among all integer partitions of n (see example below).

			Table begins:
   1:  0;
   2:  1;
   3:  2;
   4:  0,  3;
   5:  1,  4;
   6:  2,  5;
   7:  3,  6;
   8:  4,  7;
   9:  0,  5,  8;
  10:  1,  6,  9;
  11:  2,  7, 10;
  12:  3,  8, 11;
  13:  4,  9, 12;
  14:  5, 10, 13;
  15:  6, 11, 14;
  16:  0,  7, 12, 15;
  ...
For n = 4, the partitions are {4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}. The partition {2, 2} has Durfee square s = 2; for all partitions except {2, 2}, we have Durfee square with s = 1. Therefore we have two unique solutions to n - s^2 for n = 4, i.e., {0, 3}, so row 4 contains these values.

Michael De Vlieger, Table of n, a(n) for n = 1..10125 (rows 1 <= n <= 625, flattened)
Eric Weisstein's World of Mathematics, Durfee Square.

Cf. A000196, A022554, A053186, A117522.

Array[# - Reverse@ Range[Sqrt@ #]^2 &, 625] // Flatten

Let s = floor(sqrt(n));
T(n,1) = A053186(n) = n - s;
T(n,k) = T(n,1) + partial sums of 2(s - k + 1) + 1 for 2 <= k <= s + 1.
A000196(n) = Length of row n.
A022554(n) = Sum of row n.
Last term in row n = T(n, A000196(n)) = n - 1.

A270103 Array read by antidiagonals: T(n, k) is the sum of the integer part of the n-th roots of natural numbers less than k.

Original entry on oeis.org

1, 3, 1, 6, 2, 1, 10, 3, 2, 1, 15, 5, 3, 2, 1, 21, 7, 4, 3, 2, 1, 28, 9, 5, 4, 3, 2, 1, 36, 11, 6, 5, 4, 3, 2, 1, 45, 13, 7, 6, 5, 4, 3, 2, 1, 55, 16, 9, 7, 6, 5, 4, 3, 2, 1, 66, 19, 11, 8, 7, 6, 5, 4, 3, 2, 1

Offset: 1

John M. Campbell, Mar 11 2016

			The fifth entry in the second row of this array is 7, since 7 = floor(sqrt(1)) + floor(sqrt(2)) + floor(sqrt(3)) + floor(sqrt(4)) + floor(sqrt(5)).
The table array begins:
  1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
  1, 2, 3,  5,  7,  9, 11, 13, 16, 19, ...
  1, 2, 3,  4,  5,  6,  7,  9, 11, 13, ...
  1, 2, 3,  4,  5,  6,  7,  8,  9, 10, ...
  ...

R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics, 2nd Edition, Addison-Wesley, 1994, Eq. 3.27 on page 87.
D. E. Knuth, The Art of Computer Programming, Vol. 1, 3rd Edition, Addison-Wesley, 1997, Ex. 43 of section 1.2.4.

M. Griffiths, More Sums Involving the Floor Function, Math. Gaz., 86 (2002), 285-287.
Maths StackExchange, Find a formula for Sum_{k=1..n} floor(sqrt(k)), see achille hui formula.
Wikipedia, Faulhaber's formula

Rows 1 to 5: A000217, A022554, A031876, A032512, A032513.

Main diagonal and each diagonal below the main diagonal: A000027.

T[n_, k_] := (1 + k) Floor[k^(1/n)] - HarmonicNumber[Floor[k^(1/n)], -n] (* Daniel Hoying, Jun 11 2020 *)

T(n, k) = sum(j=0, k, sqrtnint(j, n)); \\ Michel Marcus, Mar 12 2016

T(n,k) = Sum_{j=0..k} floor(j^(1/n)).
T(n,k) = (1+k)*floor(k^(1/n)) - (1/(n+1))*Sum_{j=1..n+1} (1 + floor(k^(1/n)))^j*binomial(n+1, j)*Bernoulli(n+1-j).
T(n,k) = (1+k)*floor(k^(1/n)) - Sum_{j=1..floor(k^(1/n))} j^n. - Daniel Hoying, Jun 11 2020

A270825 a(n) = Sum_{i=0..n} (-1)^floor(i/2)*floor(sqrt(i)).

Original entry on oeis.org

0, 1, 0, -1, 1, 3, 1, -1, 1, 4, 1, -2, 1, 4, 1, -2, 2, 6, 2, -2, 2, 6, 2, -2, 2, 7, 2, -3, 2, 7, 2, -3, 2, 7, 2, -3, 3, 9, 3, -3, 3, 9, 3, -3, 3, 9, 3, -3, 3, 10, 3, -4, 3, 10, 3, -4, 3, 10, 3, -4, 3, 10, 3, -4, 4, 12, 4, -4, 4, 12, 4, -4, 4, 12, 4, -4, 4

Offset: 0

John M. Campbell, Mar 23 2016

			Letting [] denote the floor function, a(7) = [sqrt(0)]+[sqrt(1)]-[sqrt(2)]-[sqrt(3)]+[sqrt(4)]+[sqrt(5)]-[sqrt(6)]-[sqrt(7)] = 0+1-1-1+2+2-2-2 = -1.

Cf. A268173, A022554, A270370, A031876, A032512, A032513, A032514, A032515, A032516, A032517, A032518, A032519, A032520, A032521.

Print[Table[Sum[(-1)^(Floor[i/2])*Floor[Sqrt[i]],{i,0,n}],{n,0,100}]]

a(n)=sum(i=0,n,(-1)^(floor(i/2))*floor(sqrt(i)))

a(4m)=floor(sqrt(m)), a(4m+1)=floor(3/2*floor(sqrt(4m+1))), a(4m+2)=floor(sqrt(m)), a(4m+3)=-floor((1+sqrt(4m+3))/2).

cp's OEIS Frontend

A363499 a(n) = Sum_{k=0..n} floor(sqrt(k))^5.

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A176615 Number of edges in the graph on n vertices, labeled 1 to n, where two vertices are joined just if their labels sum to a perfect square.

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A214036 Numbers k such that floor(sqrt(1)) + floor(sqrt(2)) + floor(sqrt(3)) + ... + floor(sqrt(k)) is prime.

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A270370 a(n) = Sum_{k=0..n} (-1)^k*floor(k^(1/3)).

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A327672 a(n) = Sum_{k=0..n} ceiling(sqrt(k)).

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A262352 a(n) = Sum_{k=0..n} (-1)^k*floor(k^(1/4)).

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A331478 Irregular triangle T(n,k) = n - (s - k + 1)^2 for 1 <= k <= s, with s = floor(sqrt(n)).

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