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Row lengths for formatting A063987 as a table: The number of nonzero quadratic residues modulo a prime p equals floor(p/2), or (p-1)/2 if p is odd. The number of squares including 0 is (p+1)/2, if p is odd (rows prime(i) of A096008 formatted as a table). In fields of characteristic 2, all elements are squares. For any m > 0, floor(m/2) is the number of even positive integers less than or equal to m, so a(n) also equals the number of even positive integers less than or equal to the n-th prime. For all n > 0, A130290(n+1) = A005097(n) = A102781(n+1) = A102781(n+1) = A130291(n+1)-1 = A111333(n+1)-1 = A006254(n)-1.

From Vladimir Shevelev, Jun 18 2016: (Start)

a(1)+2 and, for n >= 2, a(n)+1 is the smallest k such that there exists 0 < k_1 < k with the condition k_1^2 == k^2 (mod prime(n)).

Indeed, for n >= 2, if prime(n) = 4*t+1 then k = 2*t+1 = a(n)+1, since (2*t+1)^2 == (2*t)^2 (mod prime(n)) and there cannot be a smaller value of k; if prime(n) = 4*t-1, then k = 2*t = a(n)+1, since (2*t)^2 == (2*t-1)^2 (mod prime(n)). (End)

a(n) is the number of pairs (a,b) such that a + b = prime(n) with 1 <= a <= b. - Nicholas Leonard, Oct 02 2022

Rows start with 1's.

Range of A002015; subset of A122986. - Reinhard Zumkeller, Mar 21 2010

Possible last three digits of n^2 (leading zeros omitted).

Range of A174452; A010461 is a subset; and also all squares less than 1000 belong to this sequence; the sequence is finite with A000993(3)=159 terms: a(159)=996 is the last term.

Range of A070442. - Reinhard Zumkeller, Apr 24 2009

Range of A159852. - Reinhard Zumkeller, Apr 24 2009

Range of A070452; a(k) + a(13-k) = 25, 1 <= k <= 12. - Reinhard Zumkeller, Apr 24 2009

Compare this sequence with A260117, Smallest triangular number that is pandigital in base n. Presumably, lim_{n->infinity} A260117(n)/A049363(n) = 1, but the same cannot be true for this sequence: the sum of the base-n digits of a number that is pandigital in base n must be 0+1+2+...+n-1 = binomial(n,2), but there are certain values of n for which no n-digit square can have a digit sum of binomial(n,2); for such values of n, a(n) must have more than n digits in base n. [E.g., the base-13 expansion of every square has a digit sum s == {0,1,4,9} (mod 12) (cf. A096008), but a square that is pandigital in base 13 and has exactly 13 digits would have a digit sum s = 78 == 6 (mod 12), so no such number exists; a 14-digit base-13 pandigital square would have each of the digits 0..12 exactly once except for one duplicated digit, which would have to be 3, 6, 7, or 10 (to yield a digit sum of 81, 84, 85, or 88, whose residues modulo 12 are 9, 0, 1, and 4, respectively). - Jon E. Schoenfield, Mar 23 2019]

The values of n for which there exists no pandigital square that is exactly n digits long (in base n) begin with 2, 3, 5, 13, 17, 21, ...; presumably, for all such values of n, a(n) is exactly n+1 base-n digits long.

In base 2, there are no 2-digit squares at all, so a(2) must have more than 2 binary digits. For n = 3, 5, 13, 17, 21, ..., there exists no square, regardless of its number of digits, whose base-n digit sum equals binomial(n,2); see A260191.

The number of squares (quadratic residues including 0) modulo a prime p (sequence A096008 with every "1" prefixed by a "0") equals 1+floor(p/2), or ceiling(p/2) = (p+1)/2 if p is odd. (In fields of characteristic 2, all elements are squares.) See A130290(n)=A130291(n)-1 for number of nonzero residues. For all n>0, A130291(n+1) = A111333(n+1) = A006254(n) = A005097(n)-1 = A102781(n+1)-1 = A102781(n+1)-1 = A130290(n+1)-1.