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Claim. Theorem 2 (p. 125) of J. A. Bondy, Large cycles in graphs, Discrete Math. 1 (1971/72), no. 2, 121--132: a graph of order $n$ and size at least $\frac12(n^2-5n+14)$ has a cycle of length $n-1$. Since $\frac12(n^2-5n+14)=\binom{n-2}2+\binom32+1$, this is the edge count of Problem 1012 at $k=1$. The paper adds that the result is best possible: a $K_{n-2}$ and a $K_3$ sharing a vertex have one edge fewer and no cycle of length $n-1$ for $n>4$. The theorem is printed without a range for $n$. No graph on $n\le3$ vertices meets the count, and for $n=4$ only $K_4$ and $K_4$ less an edge do, both with a triangle; so the implication holds for every $n\ge1$ and $f(1)=1$. The paper attributes the conjecture to Erdős at the Oxford conference of July 1969. Its § 4 (p. 128) further states that its general Conjecture 1, which is this problem's implication with $r=k+1$, holds for all $n\ge\frac12(r^2+5r+4)$, that is $f(k)\le\frac12(k+2)(k+5)$ in the problem's letters (Conjecture 1); the step for the circumference form, Conjecture 2, is asserted with "we can prove" and no written proof, so that estimate is recorded here and not claimed. The page's date is the first day of the issue's month, September 1971 (Crossref).

Covers. The case $k=1$ only, with its sharpness for $n>4$. Nothing about $k\ge2$; the estimate $f(k)\le\frac12(k+2)(k+5)$ is not covered.

Depends on. Nothing in this wiki; the paper's theorem, with Ore's theorem (the paper's Lemma 2.1), Pósa's degree condition and the theorem of Bondy's pancyclic paper, is the whole argument.

Acceptance. Refereed journal publication in Discrete Mathematics (Crossref: issue of September 1971), which is the refereed evidence; Woodall's 1972 paper (p. 749) credits the case $r=1$ to Bondy's Theorem 2. The site's label SOLVED rests on Woodall, so its credit of $f(1)=1$ to Bondy is not listed as reviewed. The proof (p. 127) is followed on the library's result page, with the inequalities of its case (b) not checked; nothing here is independently reviewed.